CommeUnJeu · L1 MPSI

Primitives --- practical techniques

⌚ ~67 min ▢ 8 blocks ✓ 31 exercises ➣ Prerequisites : Integration on a segment, Differentiability

This chapter is the practical pass at integration in this course. Given a function $f$, we learn how to find a function $F$ such that $F' = f$ --- a primitive of $f$. We work entirely at the level of primitives ; the rigorous theory of the integral $\int_a^b f(x) \, dx$ as a real (or complex) number, the fundamental theorem of calculus, and the proofs of the bounded versions of integration by parts and change of variable are the subject of chapter Integration over a segment, in semester 2. Here we treat $\int f(x) \, dx$ as a notation for « any primitive of $f$ » : a function, defined up to an additive constant. We use the lycée derivative rules and the derivative table from chapter Standard functions freely ; their formal proofs are consolidated in chapter Differentiability.
The chapter has three parts. The first part defines the primitive, presents the reading-derivatives table, and teaches the chain-rule-in-reverse pattern (« reconnaître $u' g(u)$ »). The second part introduces the two universal techniques --- integration by parts (IPP) and change of variable (COV) --- stated for primitives, with one-line proofs from the product rule and the chain rule. The third part presents three closed-form recipes : primitives of $e^{ax} \cos(bx)$ via the complex exponential, primitives of $1/(ax^2 + bx + c)$ when the discriminant is negative, and partial-fraction decomposition in simple cases.
Three reflexes the reader should leave with : (i) verify every primitive by differentiation --- a one-line sanity check that catches most errors ; (ii) IPP and COV are recipes, not theorems with $\int_a^b$ --- they say « if $H$ primitive of $uv'$, then $uv - H$ primitive of $u'v$ » and « $F$ primitive of $f$ implies $F \circ \varphi$ primitive of $(f \circ \varphi) \varphi'$ » ; the bounded versions live in chapter Integration over a segment ; (iii) every primitive lives on an interval : « find a primitive of $1/x$ » means « on $\mathbb R_+^*$ » or « on $\mathbb R_-^*$ », not on $\mathbb R^*$, because uniqueness à constante additive près only holds on a single interval.

I Primitives --- definitions$\virgule$ table$\virgule$ identification

I.1 Definition of a primitive

A primitive of $f$ on an interval $I$ is a function $F$ that is differentiable on $I$ with $F' = f$. There is never the primitive --- only a primitive : the operation $F \mapsto F'$ kills constants, so primitives come in families, all of which differ by an additive constant. This is the « uniqueness à constante additive près » that drives the choice of an interval $I$ as the natural domain.

Definition — Primitive of a function

Let $I$ be an interval of $\mathbb R$ and $f : I \to \mathbb{K}$ a function, where $\mathbb{K} = \mathbb R$ or $\mathbb C$. A primitive of $f$ on $I$ is a function $F : I \to \mathbb{K}$ such that $F$ is differentiable on $I$ and $F'(x) = f(x)$ for every $x \in I$. We sometimes write $F(x) = \int f(x) \, dx$ or $F = \int f$.

Notation $\int f \mathrm{d}x$ vs definite integral

Throughout this chapter, $\int f(x) \, dx$ denotes a primitive of $f$ --- a function, defined up to an additive constant. The rigorous definition of the integral $\int_a^b f(x) \, dx$ as a number (the « area under the curve », defined via Riemann sums) is the subject of chapter Integration over a segment (semester 2). Here, $\int f \, dx$ is just notation.

Proposition — Uniqueness à constante additive près

Let $f : I \to \mathbb{K}$ admit a primitive $F$ on the interval $I$. Then the primitives of $f$ on $I$ are exactly the functions of the form $F + c$ for $c \in \mathbb{K}$. In particular, two primitives of $f$ on $I$ differ by a constant.

Proof

Let $G : I \to \mathbb{K}$ be a primitive of $f$ on $I$. Then $(G - F)'(x) = G'(x) - F'(x) = f(x) - f(x) = 0$ for every $x \in I$. We use the standard fact that a real-valued function whose derivative is zero on an interval is constant on that interval. This fact will be proved rigorously in chapter Differentiability (chapter 16) as a consequence of the mean value theorem ; at this stage we cite it from the lycée. For $\mathbb{K} = \mathbb C$, we apply this real-valued result componentwise to the real and imaginary parts of $G - F$ to conclude that $G - F$ is constant. So $G - F$ is a constant function on $I$, say $G(x) - F(x) = c$ for every $x \in I$, i.e. $G = F + c$. Conversely, for every $c \in \mathbb{K}$, the function $F + c$ has derivative $F' = f$, hence is a primitive of $f$.

Proposition — Linearity of primitives

Let $F$ be a primitive of $f$ and $G$ a primitive of $g$ on an interval $I$. For all $\lambda, \mu \in \mathbb{K}$, the function $\lambda F + \mu G$ is a primitive of $\lambda f + \mu g$ on $I$.

Proof

By the linearity of the derivative, $(\lambda F + \mu G)'(x) = \lambda F'(x) + \mu G'(x) = \lambda f(x) + \mu g(x)$ for every $x \in I$.

Method — Verifying that $F$ is a primitive of $f$

Whenever you claim « $F$ is a primitive of $f$ », differentiate $F$ and check $F' = f$ on the announced interval. This one-line sanity check catches most errors in primitive computations and is expected in every written answer.

Example — A polynomial primitive

The function $F : x \mapsto \dfrac{x^3}{3} + x^2 - 5 x + 7$ is a primitive of $f : x \mapsto x^2 + 2 x - 5$ on $\mathbb R$. Verification : $F'(x) = x^2 + 2 x - 5 = f(x)$. Any other primitive of $f$ on $\mathbb R$ is of the form $F + c$ for some real $c$.

Example — A non-elementary case

The function $x \mapsto e^{-x^2}$ admits primitives on $\mathbb R$ (every continuous function does ; this will be made rigorous in chapter Integration over a segment), but no primitive of $e^{-x^2}$ can be expressed using the elementary functions $\exp$, $\log$, $\sin$, $\arctan$, etc. This fact is admitted here and lies beyond the program. We can still talk about $\int e^{-x^2} \, dx$ as a formal primitive ; we just cannot reduce it to a known function.

Skills to practice

Verifying a primitive and recovering it from $F'$

I.2 Common primitives --- the table

Every entry in the derivatives table from chapter Standard functions, read in reverse, gives a primitive. We collect the most-used entries below ; the student is expected to know this table by heart.

Proposition — Table of usual primitives

Each line below gives a primitive on the indicated interval (where $a$ is a real parameter and $c \in \mathbb{K}$ is an arbitrary constant ; $\mathbb{K} = \mathbb R$ or $\mathbb C$) :

$f(x) = e^x$ on $\mathbb R$ : primitive $F(x) = e^x + c$.
$f(x) = \ln x$ on $\mathbb R_+^*$ : primitive $F(x) = x \ln x - x + c$.
$f(x) = 1/x$ on any interval contained in $\mathbb R^*$ : primitive $F(x) = \ln |x| + c$.
$f(x) = x^a$ for $a \in \mathbb R \setminus \{-1\}$, on $\mathbb R_+^*$ : primitive $F(x) = \dfrac{x^{a+1}}{a + 1} + c$.
$f(x) = \cos x$ on $\mathbb R$ : primitive $F(x) = \sin x + c$.
$f(x) = \sin x$ on $\mathbb R$ : primitive $F(x) = -\cos x + c$.
$f(x) = \tan x$ on any interval not meeting $\pi/2 + \pi \mathbb Z$ : primitive $F(x) = -\ln |\cos x| + c$.
$f(x) = \cosh x$ on $\mathbb R$ : primitive $F(x) = \sinh x + c$.
$f(x) = \sinh x$ on $\mathbb R$ : primitive $F(x) = \cosh x + c$.
$f(x) = \tanh x$ on $\mathbb R$ : primitive $F(x) = \ln(\cosh x) + c$.
$f(x) = \dfrac{1}{1 + x^2}$ on $\mathbb R$ : primitive $F(x) = \arctan x + c$.
$f(x) = \dfrac{1}{\sqrt{1 - x^2}}$ on $]-1, 1[$ : primitive $F(x) = \arcsin x + c$.

Proof

Each line is verified by differentiating the proposed primitive and recovering $f$. We illustrate three :

$\bigl(\ln |x|\bigr)' = 1/x$ on $\mathbb R^*$, using the chain rule on $\ln |x| = \ln x$ (for $x > 0$) or $\ln(-x)$ (for $x < 0$).
$\bigl(x \ln x - x\bigr)' = \ln x + x \cdot \dfrac{1}{x} - 1 = \ln x$ on $\mathbb R_+^*$, by the product rule.
$\bigl(\arctan x\bigr)' = \dfrac{1}{1 + x^2}$ on $\mathbb R$, by the derivative of $\arctan$ from chapter Standard functions.

The remaining lines follow identically from the derivative table of that chapter.

Method — Reading the derivatives table in reverse

When asked for a primitive of $f$, first scan the usual derivatives table from chapter Standard functions : is $f$ already in the « right column » of some standard derivative ? If yes, the primitive is the « left column ». This is the fastest route whenever it works.

Example — Reading the table directly

A primitive of $f(x) = 3 x^2 - 2 \cos x + \dfrac{1}{x}$ on $\mathbb R_+^*$ is $$ F(x) \ = \ x^3 - 2 \sin x + \ln x + c, \qquad c \in \mathbb R. $$ Verification : $F'(x) = 3 x^2 - 2 \cos x + 1/x = f(x)$ on $\mathbb R_+^*$.

Example — Combining several entries

A primitive of $f(x) = x^{-3} + e^x + \dfrac{1}{1 + x^2}$ on $\mathbb R_+^*$ is $$ F(x) \ = \ -\dfrac{1}{2 x^2} + e^x + \arctan x + c. $$ (For the $x^{-3}$ piece, use the line $x^a \mapsto x^{a+1}/(a+1)$ with $a = -3$, giving $x^{-2}/(-2) = -1/(2 x^2)$.)

Skills to practice

Reading the primitive table

I.3 Identification --- recognizing $u' \ g(u)$

The chain rule of differentiation reads $(g \circ u)' = u' \cdot (g' \circ u)$. Reading it backwards : if $G$ is a primitive of $g$, then $G \circ u$ is a primitive of $u' \cdot (g \circ u)$. This pattern-matching reflex --- spot the composition $u'(x) \cdot g(u(x))$ in the integrand, then write $G(u(x))$ for the primitive --- is the workhorse of primitive-finding.

Proposition — Identification --- pattern $u' \ g(u)$

Let $I, J$ be intervals of $\mathbb R$, $u : I \to J$ differentiable on $I$, and $G : J \to \mathbb{K}$ a primitive of a function $g : J \to \mathbb{K}$ on $J$. Then $G \circ u$ is a primitive of $u' \cdot (g \circ u)$ on $I$. The most useful instances of this pattern are :

$u' \, e^u$ admits $e^u$ as primitive.
$\dfrac{u'}{u}$ admits $\ln |u|$ as primitive (on any interval where $u \ne 0$).
$u' \, u^a$ for $a \ne -1$ admits $\dfrac{u^{a+1}}{a+1}$ as primitive (on any interval where $u^a$ is defined).
$u' \cos u$ admits $\sin u$ as primitive ; $u' \sin u$ admits $-\cos u$.
$\dfrac{u'}{1 + u^2}$ admits $\arctan u$ as primitive.
$\dfrac{u'}{\sqrt{1 - u^2}}$ admits $\arcsin u$ as primitive (on any interval where $|u| < 1$).

Proof

By the chain rule, $(G \circ u)'(x) = G'(u(x)) \cdot u'(x) = g(u(x)) \cdot u'(x) = u'(x) \, g(u(x))$. Hence $G \circ u$ is a primitive of $u' \cdot (g \circ u)$ on $I$. The listed instances are obtained by taking $g$ to be respectively $\exp$, $\mathrm{inv}$, $x \mapsto x^a$, $\cos$, $\sin$, $x \mapsto 1/(1 + x^2)$, $x \mapsto 1/\sqrt{1 - x^2}$, and reading the corresponding primitive $G$ from the table of usual primitives.

Proposition — Affine substitution

Let $F$ be a primitive of $f$ on an interval $J$. For every $a \in \mathbb R^*$ and $b \in \mathbb R$, the function $x \mapsto \dfrac{1}{a} F(a x + b)$ is a primitive of $x \mapsto f(a x + b)$ on every interval $I$ such that $\{a x + b \mid x \in I\} \subset J$.

Proof

By the chain rule with $u(x) = a x + b$, $u'(x) = a$ : $$ \left( \dfrac{1}{a} F(a x + b) \right)' \ = \ \dfrac{1}{a} \cdot a \cdot F'(a x + b) \ = \ F'(a x + b) \ = \ f(a x + b). $$

Method — Identify the $u' \ g(u)$ pattern

To find a primitive of a complicated integrand, first try pattern matching :

Scan for a factor that could be $u'$ (the derivative of an « inside » expression).
Identify what « inside » expression $u$ that derivative belongs to.
Check that the rest of the integrand is $g(u)$ for some known $g$ in the table.
Write the primitive as $G(u(x))$ where $G$ is a primitive of $g$, and verify by differentiation.

The affine-substitution case $f(a x + b)$ is the simplest instance : $u(x) = a x + b$, $u' = a$ constant, so $\int f(a x + b) \, dx = \dfrac{1}{a} F(a x + b)$ where $F$ is a primitive of $f$.

Example — Logarithmic pattern

On $\mathbb R$, a primitive of $f(x) = \dfrac{x}{1 + x^2}$ is found by spotting that the numerator $x$ is (up to a factor $1/2$) the derivative of the denominator $1 + x^2$. Setting $u(x) = 1 + x^2$, $u'(x) = 2 x$, we recognize $\dfrac{1}{2} \cdot \dfrac{u'}{u}$, whose primitive is $\dfrac{1}{2} \ln |u|$ : $$ \int \dfrac{x}{1 + x^2} \, dx \ = \ \dfrac{1}{2} \ln(1 + x^2) + c. $$ (The absolute value drops because $1 + x^2 > 0$.) Verification : $\bigl(\dfrac{1}{2} \ln(1 + x^2)\bigr)' = \dfrac{1}{2} \cdot \dfrac{2 x}{1 + x^2} = \dfrac{x}{1 + x^2}$. $\checkmark$

Example — Arctan pattern

On $\mathbb R$, a primitive of $f(x) = \dfrac{e^x}{1 + e^{2 x}}$ is found by setting $u(x) = e^x$, so $u'(x) = e^x$ and $u^2 = e^{2 x}$ ; the integrand is $\dfrac{u'}{1 + u^2}$, primitive $\arctan u$ : $$ \int \dfrac{e^x}{1 + e^{2 x}} \, dx \ = \ \arctan(e^x) + c. $$ Verification : $\bigl(\arctan(e^x)\bigr)' = \dfrac{e^x}{1 + e^{2 x}}$. $\checkmark$

Example — Affine substitution

On $\mathbb R$, a primitive of $f(x) = \cos(2 x + \pi/3)$ is found via the affine-substitution rule with $a = 2$, $b = \pi/3$, $F = \sin$ : $$ \int \cos(2 x + \pi/3) \, dx \ = \ \dfrac{1}{2} \sin(2 x + \pi/3) + c. $$ Verification : $\bigl(\dfrac{1}{2} \sin(2 x + \pi/3)\bigr)' = \dfrac{1}{2} \cdot 2 \cos(2 x + \pi/3) = \cos(2 x + \pi/3)$. $\checkmark$

Skills to practice

Identifying $u' \ g(u)$ patterns

II Two integration techniques

II.1 Integration by parts

The product rule of differentiation says $(u v)' = u' v + u v'$. Read backwards, this gives the integration by parts recipe : to find a primitive of $u' v$, one transfers the derivation from $u$ to $v$, paying the price $u v$ along the way. The art is to choose $u'$ and $v$ so that the new integrand $u v'$ is simpler than the original $u' v$.

Proposition — Integration by parts for primitives --- IPP

Let $I$ be an interval of $\mathbb R$ and $u, v : I \to \mathbb{K}$ be of class $C^1$ on $I$. If $H : I \to \mathbb{K}$ is a primitive of $u v'$ on $I$, then $u v - H$ is a primitive of $u' v$ on $I$. We write this informally as $$ \int u'(x) v(x) \, dx \ = \ u(x) v(x) - \int u(x) v'(x) \, dx. $$

Proof

By the product rule and the hypothesis $H' = u v'$, $$ \begin{aligned} (u v - H)'(x) \ &= \ (u v)'(x) - H'(x) && \text{(linearity)} \\ &= \ u'(x) v(x) + u(x) v'(x) - u(x) v'(x) && \text{(product rule and $H' = u v'$)} \\ &= \ u'(x) v(x). \end{aligned} $$ Hence $u v - H$ is a primitive of $u' v$ on $I$.

Bounded version

The bounded version $\int_a^b u'(x) v(x) \, dx = \bigl[u v\bigr]_a^b - \int_a^b u(x) v'(x) \, dx$ requires the rigorous theory of the definite integral and the fundamental theorem of calculus, both of which are treated in chapter Integration over a segment (semester 2). Here we work with primitives only.

Method — Apply IPP

To find a primitive of a product, identify one factor as $u'$ and the other as $v$ :

Choose a factor $u'$ whose primitive $u$ is easy to compute.
Choose a factor $v$ whose derivative $v'$ is simpler than $v$ itself.
Write the IPP identity $\int u' v = u v - \int u v'$.
Check that $\int u v'$ is indeed simpler ; if not, swap the choices of $u'$ and $v$.

Classical signpost cases : $\int P(x) e^{a x} \, dx$ (take $u' = e^{a x}$), $\int P(x) \cos(a x) \, dx$ (take $u' = \cos(a x)$), $\int P(x) \ln x \, dx$ (take $v = \ln x$), $\int \arctan x \, dx$ (« invisible 1 » trick : $u' = 1$, $v = \arctan x$), $\int \ln x \, dx$ (same trick).

Example — Primitive of $x \ e^x$

On $\mathbb R$, take $u'(x) = e^x$ (primitive $u(x) = e^x$) and $v(x) = x$ (derivative $v'(x) = 1$). IPP gives $$ \int x \, e^x \, dx \ = \ x \, e^x - \int e^x \cdot 1 \, dx \ = \ x \, e^x - e^x + c \ = \ (x - 1) e^x + c. $$ Verification : $\bigl((x - 1) e^x\bigr)' = e^x + (x - 1) e^x = x \, e^x$. $\checkmark$

Example — Primitive of $\ln x$ via the « invisible 1 »

On $\mathbb R_+^*$, write $\ln x = 1 \cdot \ln x$ and apply IPP with $u'(x) = 1$ (primitive $u(x) = x$) and $v(x) = \ln x$ (derivative $v'(x) = 1/x$) : $$ \int \ln x \, dx \ = \ x \ln x - \int x \cdot \dfrac{1}{x} \, dx \ = \ x \ln x - x + c. $$ Verification : $\bigl(x \ln x - x\bigr)' = \ln x + 1 - 1 = \ln x$. $\checkmark$ This recovers the table entry for $\ln x$ from the reading-derivatives subsection above.

Skills to practice

Applying integration by parts

II.2 Change of variable

The chain rule says $(F \circ \varphi)'(x) = (f \circ \varphi)(x) \cdot \varphi'(x)$ when $F' = f$. Read backwards, this is the change of variable recipe : when the integrand contains both a function and its derivative, set $u = \varphi(t)$ and the integrand becomes $f(u) \, du$ --- for which we already know how to find a primitive.

Proposition — Change of variable for primitives --- COV

Let $I, J$ be intervals of $\mathbb R$, $\varphi : I \to J$ a differentiable function, $f : J \to \mathbb{K}$, and $F : J \to \mathbb{K}$ a primitive of $f$ on $J$. Then $F \circ \varphi$ is a primitive of $(f \circ \varphi) \cdot \varphi'$ on $I$. Informally, setting $u = \varphi(t)$ with $du = \varphi'(t) \, dt$ : $$ \int f(\varphi(t)) \, \varphi'(t) \, dt \ = \ \int f(u) \, du \ = \ F(u) + c \ = \ F(\varphi(t)) + c. $$

Proof

By the chain rule applied to $F \circ \varphi$ on $I$ : $$ (F \circ \varphi)'(t) \ = \ F'(\varphi(t)) \cdot \varphi'(t) \ = \ f(\varphi(t)) \cdot \varphi'(t). $$ Hence $F \circ \varphi$ is a primitive of $(f \circ \varphi) \cdot \varphi'$ on $I$.

Bounded version

The bounded version $\int_a^b f(\varphi(t)) \, \varphi'(t) \, dt = \int_{\varphi(a)}^{\varphi(b)} f(u) \, du$ requires the rigorous theory of the definite integral and is treated in chapter Integration over a segment.

Method — Apply COV

To find a primitive of an integrand of the form $f(\varphi(t)) \cdot \varphi'(t)$ :

Set $u = \varphi(t)$.
Write the differential $du = \varphi'(t) \, dt$.
Rewrite the integrand entirely in terms of $u$ : it should reduce to $f(u) \, du$.
Find a primitive $F$ of $f$ in the new variable.
Substitute back : the primitive of the original integrand is $F(\varphi(t))$.

Classical signpost cases : $\int f(\sqrt x) \, dx$ via $u = \sqrt x$ ; $\int f(\sin x) \cos x \, dx$ via $u = \sin x$ ; $\int f(\ln x)/x \, dx$ via $u = \ln x$ ; $\int f(e^x) e^x \, dx$ via $u = e^x$.

Example — Primitive of $1/(x \ln x)$

On any interval contained in $]0, 1[$ or in $]1, +\infty[$ (so that $\ln x$ does not vanish), set $u = \ln x$, $du = dx/x$. The integrand $\dfrac{1}{x \ln x} dx = \dfrac{du}{u}$, primitive $\ln |u| + c$. Substituting back : $$ \int \dfrac{dx}{x \ln x} \ = \ \ln |\ln x| + c. $$ Verification : $\bigl(\ln |\ln x|\bigr)' = \dfrac{1/x}{\ln x} = \dfrac{1}{x \ln x}$. $\checkmark$

Example — Primitive of square root of $1 - x^2$

On $]-1, 1[$, we apply an inverse substitution : set $x = \sin \theta$ for $\theta \in {]-\pi/2, \pi/2[}$. This is justified because $\arcsin : {]-1, 1[} \to {]-\pi/2, \pi/2[}$ is a $C^1$ bijection (chapter Standard functions), so $\theta = \arcsin x$ is a legitimate change of variable on this interval ; the final formula is verified by differentiation below. We get $dx = \cos \theta \, d\theta$ and $\sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = |\cos \theta| = \cos \theta$ (positive on $]-\pi/2, \pi/2[$). The integrand becomes $$ \sqrt{1 - x^2} \, dx \ = \ \cos \theta \cdot \cos \theta \, d\theta \ = \ \cos^2 \theta \, d\theta \ = \ \dfrac{1 + \cos(2 \theta)}{2} \, d\theta $$ using the linearisation formulas from chapter Trigonometry. A primitive in $\theta$ is $\dfrac{\theta}{2} + \dfrac{\sin(2 \theta)}{4} = \dfrac{\theta + \sin \theta \cos \theta}{2}$. Substituting back $\theta = \arcsin x$, $\sin \theta = x$, $\cos \theta = \sqrt{1 - x^2}$ : $$ \int \sqrt{1 - x^2} \, dx \ = \ \dfrac{\arcsin x + x \sqrt{1 - x^2}}{2} + c. $$ Verification by differentiation : with $F(x) = \dfrac{\arcsin x + x \sqrt{1 - x^2}}{2}$, $$ F'(x) \ = \ \dfrac{1}{2}\!\left[\dfrac{1}{\sqrt{1 - x^2}} + \sqrt{1 - x^2} + x \cdot \dfrac{-x}{\sqrt{1 - x^2}}\right] \ = \ \dfrac{1}{2}\!\left[\dfrac{1 + (1 - x^2) - x^2}{\sqrt{1 - x^2}}\right] \ = \ \dfrac{2 - 2 x^2}{2 \sqrt{1 - x^2}} \ = \ \sqrt{1 - x^2}. \ \checkmark $$

Skills to practice

Applying change of variable

III Closed-form techniques for usual families

III.1 Primitives of $e^{ax} \cos(bx)$ and $e^{ax} \sin(bx)$ --- the complex method

The cleanest way to find a primitive of $e^{a x} \cos(b x)$ is via the complex exponential : $e^{a x} \cos(b x) = \mathrm{Re}\bigl(e^{(a + i b) x}\bigr)$. Since the complex exponential admits an explicit primitive just like the real one --- $\int e^{c x} \, dx = e^{c x}/c$ for $c \ne 0$ --- the closed form follows by taking real parts. This route avoids the double-IPP detour and uses the complex-exponential calculus from chapter Complex numbers.

Proposition — Closed-form primitives of $e^{ax} \cos(bx)$ and $e^{ax} \sin(bx)$

Let $a, b \in \mathbb R$ not both zero. On $\mathbb R$, a primitive of $f(x) = e^{a x} \cos(b x)$ is $$ F(x) \ = \ \dfrac{a \cos(b x) + b \sin(b x)}{a^2 + b^2} \cdot e^{a x}, $$ and a primitive of $g(x) = e^{a x} \sin(b x)$ is $$ G(x) \ = \ \dfrac{a \sin(b x) - b \cos(b x)}{a^2 + b^2} \cdot e^{a x}. $$

Proof

Set $c = a + i b$ ; then $c \ne 0$ since $a$ and $b$ are not both zero. By the derivative of the complex exponential (chapter Complex numbers), $\dfrac{d}{dx}\bigl(e^{c x}\bigr) = c \, e^{c x}$. Hence $\dfrac{e^{c x}}{c}$ is a primitive of $e^{c x}$. Now decompose into real and imaginary parts using $1/c = \overline{c}/|c|^2 = (a - i b)/(a^2 + b^2)$ : $$ \begin{aligned} \dfrac{e^{c x}}{c} \ &= \ \dfrac{a - i b}{a^2 + b^2} \cdot e^{a x} (\cos(b x) + i \sin(b x)) && \text{(definition of $c$ and Euler)} \\ &= \ \dfrac{e^{a x}}{a^2 + b^2} \bigl[ (a - i b)(\cos(b x) + i \sin(b x)) \bigr] && \text{(factor $e^{ax}$)} \\ &= \ \dfrac{e^{a x}}{a^2 + b^2} \bigl[ (a \cos(b x) + b \sin(b x)) + i (a \sin(b x) - b \cos(b x)) \bigr] && \text{(expand)}. \end{aligned} $$ Taking the real part gives $F(x)$ (a primitive of $\mathrm{Re}(e^{c x}) = e^{a x} \cos(b x)$), and taking the imaginary part gives $G(x)$ (a primitive of $\mathrm{Im}(e^{c x}) = e^{a x} \sin(b x)$).

Method — Primitives via the complex method

For an integrand of the form $e^{a x} \cos(b x)$ or $e^{a x} \sin(b x)$ :

Write $e^{a x} \cos(b x) = \mathrm{Re}\bigl(e^{(a + i b) x}\bigr)$ (resp.\ $\mathrm{Im}$ for the $\sin$ case).
Primitivate the complex exponential : $\int e^{(a + i b) x} \, dx = \dfrac{e^{(a + i b) x}}{a + i b}$.
Take the real (resp.\ imaginary) part of the result.

A double-IPP route exists (two integrations by parts on $e^{a x} \cos(b x)$ recover the same primitive) but is significantly more tedious.

Example — Primitive of $e^x \cos(2x)$

Take $a = 1$, $b = 2$, so $a^2 + b^2 = 5$. The Proposition gives $$ \int e^x \cos(2 x) \, dx \ = \ \dfrac{\cos(2 x) + 2 \sin(2 x)}{5} \cdot e^x + c. $$ Verification : differentiating the candidate primitive gives $$ \dfrac{d}{dx}\!\left[\dfrac{e^x (\cos(2 x) + 2 \sin(2 x))}{5}\right] \ = \ \dfrac{e^x (\cos(2 x) + 2 \sin(2 x)) + e^x (-2 \sin(2 x) + 4 \cos(2 x))}{5} \ = \ \dfrac{e^x \cdot 5 \cos(2 x)}{5} \ = \ e^x \cos(2 x). \ \checkmark $$

Example — Primitive of $e^{-x} \sin x$

Take $a = -1$, $b = 1$, so $a^2 + b^2 = 2$. The Proposition gives $$ \int e^{-x} \sin x \, dx \ = \ \dfrac{-\sin x - \cos x}{2} \cdot e^{-x} + c \ = \ -\dfrac{e^{-x} (\sin x + \cos x)}{2} + c. $$ A direct differentiation verifies the result.

Skills to practice

Computing primitives by the complex method

III.2 Primitives of $1/(Ax^2 + Bx + C)$ with negative discriminant

When the denominator $A x^2 + B x + C$ has no real roots (discriminant $B^2 - 4 A C < 0$), one completes the square to write it as $A \bigl[(x - \alpha)^2 + \beta^2\bigr]$. The integrand then matches the $\arctan$ pattern $1/(u^2 + \beta^2)$. (We use capital letters here for the trinomial coefficients to avoid a clash with the parameter $a$ in the boxed identity below.)

Proposition — Boxed identity --- $1/(x^2 + a^2)$

For every $a > 0$, the function $x \mapsto \dfrac{1}{a} \arctan \dfrac{x}{a}$ is a primitive of $x \mapsto \dfrac{1}{x^2 + a^2}$ on $\mathbb R$.

Proof

By the chain rule applied to $\arctan(x/a)$ : $$ \dfrac{d}{dx}\!\left(\dfrac{1}{a} \arctan \dfrac{x}{a}\right) \ = \ \dfrac{1}{a} \cdot \dfrac{1/a}{1 + (x/a)^2} \ = \ \dfrac{1}{a^2 + x^2}. $$

Method — Complete the square

To find a primitive of $1/(A x^2 + B x + C)$ when the discriminant is negative :

Factor out the leading coefficient : $A x^2 + B x + C = A (x^2 + (B/A) x + C/A)$.
Complete the square : $x^2 + (B/A) x + C/A = (x + B/(2A))^2 + (C/A - B^2/(4 A^2))$. Set $\alpha = -B/(2A)$ and $\beta^2 = C/A - B^2/(4 A^2)$ ; we choose $\beta > 0$ (possible because the discriminant is negative, so $\beta^2 > 0$).
Apply the boxed identity by an affine substitution $u = x - \alpha$ : primitive $= \dfrac{1}{A \beta} \arctan \dfrac{x - \alpha}{\beta}$.

Example — Primitive of $1/(x^2 + 2x + 5)$

Complete the square : $x^2 + 2 x + 5 = (x + 1)^2 + 4$. Apply the boxed identity with $a = 2$ and a shift by $1$ : $$ \int \dfrac{dx}{x^2 + 2 x + 5} \ = \ \dfrac{1}{2} \arctan \dfrac{x + 1}{2} + c. $$ Verification : $\dfrac{d}{dx}\!\bigl[\dfrac{1}{2} \arctan \dfrac{x+1}{2}\bigr] = \dfrac{1}{2} \cdot \dfrac{1/2}{1 + ((x+1)/2)^2} = \dfrac{1}{4 + (x+1)^2} = \dfrac{1}{x^2 + 2 x + 5}$. $\checkmark$

Example — Primitive of $1/(2x^2 + 4x + 3)$

Factor out $2$ : $2 x^2 + 4 x + 3 = 2 \bigl[(x + 1)^2 + 1/2\bigr]$. With $\alpha = -1$ and $\beta = 1/\sqrt 2$ : $$ \int \dfrac{dx}{2 x^2 + 4 x + 3} \ = \ \dfrac{1}{2 \cdot (1/\sqrt 2)} \arctan \dfrac{x + 1}{1/\sqrt 2} + c \ = \ \dfrac{1}{\sqrt 2} \arctan\!\bigl(\sqrt 2 (x + 1)\bigr) + c. $$ A direct differentiation verifies the result.

Skills to practice

Completing the square and applying $\arctan$

III.3 Decomposition into partial fractions --- simple cases

A rational function $P(x)/Q(x)$ can often be split into a sum of simpler fractions --- one per linear factor of $Q$, one per irreducible quadratic factor --- each of which already admits a known primitive from the table or one of the recipes above. The full theory of this decomposition into partial fractions (DES, « décomposition en éléments simples ») is the subject of chapter Polynomial arithmetic (chapter 23). Here we cover four simple cases that suffice for primitive computations in this chapter and most of analysis.

Proposition — Decomposition into partial fractions --- four simple cases

The following simple decompositions are verified by bringing the right-hand side to a common denominator. The general theory explaining why such decompositions always exist in the appropriate setting is deferred to chapter Polynomial arithmetic.

Type 1. For $a \ne b$ in $\mathbb R$, $\dfrac{1}{(x - a)(x - b)} = \dfrac{1}{a - b} \cdot \dfrac{1}{x - a} + \dfrac{1}{b - a} \cdot \dfrac{1}{x - b}$.
Type 2. The fraction $\dfrac{1}{(x - a)^2}$ is already in simple form ; primitive $-\dfrac{1}{x - a}$.
Type 3. For $p, q \in \mathbb R$ with $p^2 - 4 q < 0$, $\dfrac{\alpha x + \beta}{x^2 + p x + q}$ splits as $\dfrac{\alpha}{2} \cdot \dfrac{2 x + p}{x^2 + p x + q} + \dfrac{2 \beta - \alpha p}{2} \cdot \dfrac{1}{x^2 + p x + q}$. The first piece is a $u'/u$ pattern (giving $\ln$) ; the second piece is the canonical-form arctan recipe from the previous subsection.
Type 4. The fraction $\dfrac{1}{x (x^2 + 1)}$ decomposes as $\dfrac{1}{x} - \dfrac{x}{x^2 + 1}$, and primitives follow term by term. Analogous decompositions hold for $1/((x - a)(x^2 + p x + q))$ with a negative-discriminant quadratic.

Reference forward

The general theory of partial-fraction decomposition --- Euclidean division of polynomials to extract the integer part, irreducible factorisations over $\mathbb R$ and $\mathbb C$, the systematic techniques (« cover-up », limit at $+\infty$, evaluation at a point, identification) for computing coefficients --- is the subject of chapter Polynomial arithmetic (chapter 23). The four cases above are stated as recipes that the student can apply by inspection.

Method — Cover-up technique for linear factors

To find the coefficient of $\dfrac{1}{x - \lambda}$ in a Type 1 or Type 4 decomposition :

Multiply both sides of the decomposition by $(x - \lambda)$.
Evaluate the result at $x = \lambda$ ; the right-hand side reduces to the single coefficient sought.
Read off the coefficient.

This « cover-up » technique is the fastest route for Type 1 and Type 4. For Type 3 (the $(\alpha x + \beta)/(x^2 + p x + q)$ piece), use the split described in the Proposition.

Example — Type 1 --- $1/((x-1)(x+2))$

On any interval contained in $\mathbb R \setminus \{-2, 1\}$, the Type 1 decomposition with $a = 1$, $b = -2$ gives $a - b = 3$ : $$ \dfrac{1}{(x - 1)(x + 2)} \ = \ \dfrac{1/3}{x - 1} - \dfrac{1/3}{x + 2}. $$ Each piece is $u'/u$, with primitives $\dfrac{1}{3} \ln |x - 1|$ and $-\dfrac{1}{3} \ln|x + 2|$. A primitive of the original is $$ \int \dfrac{dx}{(x - 1)(x + 2)} \ = \ \dfrac{1}{3} \ln \left| \dfrac{x - 1}{x + 2} \right| + c. $$ A direct differentiation verifies the result.

Example — Type 3 --- split into log plus arctan

The denominator $x^2 + 2 x + 5$ has discriminant $4 - 20 < 0$, so never vanishes ; work on $\mathbb R$. The numerator is $x$ ; we engineer it so the $u' = 2 x + 2$ pattern appears : $$ x \ = \ \dfrac{1}{2} (2 x + 2) - 1, \qquad \dfrac{x}{x^2 + 2 x + 5} \ = \ \dfrac{1}{2} \cdot \dfrac{2 x + 2}{x^2 + 2 x + 5} - \dfrac{1}{x^2 + 2 x + 5}. $$ First piece ($u'/u$ pattern with $u = x^2 + 2 x + 5$, $u' = 2 x + 2$) : primitive $\dfrac{1}{2} \ln(x^2 + 2 x + 5)$. Second piece (canonical form from the previous subsection) : $x^2 + 2 x + 5 = (x + 1)^2 + 4$, primitive $\dfrac{1}{2} \arctan \dfrac{x + 1}{2}$.
Combining : $$ \int \dfrac{x}{x^2 + 2 x + 5} \, dx \ = \ \dfrac{1}{2} \ln(x^2 + 2 x + 5) - \dfrac{1}{2} \arctan \dfrac{x + 1}{2} + c. $$ A direct differentiation verifies the result.

Example — Type 4 --- $1/(x(x^2+1))$ by cover-up

On $\mathbb R_+^*$ (or on $\mathbb R_-^*$), look for a decomposition $\dfrac{1}{x (x^2 + 1)} = \dfrac{a}{x} + \dfrac{b x + c}{x^2 + 1}$ with $a, b, c \in \mathbb R$. Multiply both sides by $x$ and evaluate at $x = 0$ : $1/(0^2 + 1) = a$, so $a = 1$. Then $\dfrac{1}{x (x^2 + 1)} - \dfrac{1}{x} = \dfrac{1 - (x^2 + 1)}{x (x^2 + 1)} = \dfrac{-x^2}{x (x^2 + 1)} = -\dfrac{x}{x^2 + 1}$, identifying $b = -1$, $c = 0$. So $$ \dfrac{1}{x (x^2 + 1)} \ = \ \dfrac{1}{x} - \dfrac{x}{x^2 + 1}. $$ Each piece admits an explicit primitive : on any interval contained in $\mathbb R_+^*$ or $\mathbb R_-^*$, $$ \int \dfrac{dx}{x (x^2 + 1)} \ = \ \ln |x| - \dfrac{1}{2} \ln(1 + x^2) + c \ = \ \ln\!\left(\dfrac{|x|}{\sqrt{1 + x^2}}\right) + c. $$ A direct differentiation verifies the result.

Skills to practice

Partial-fraction decomposition and primitivation