Integration by Parts, Partial Fractions and Special Integrals notes: ISC Class 12 Mathematics

1. Integration by parts

From the product rule, $\displaystyle\int u\,dv = uv - \int v\,du$ . Choose $u$ in the order inverse trigonometric, logarithmic, algebraic, trigonometric, exponential. For $\int \log x\,dx$ and $\int \tan^{-1} x\,dx$ , take $dv = dx$ . When the original integral reappears (as with $e^{ax}\sin bx$ ), apply parts twice with the same kind of choice and solve for it.

Pick $u$ from the higher row; the other factor is $dv$.
Priority for $u$	Type	Example factor
1	Inverse trigonometric	$\tan^{-1} x$ , $\sin^{-1} x$
2	Logarithmic	$\log x$ , $\log 3x$
3	Algebraic	$x$ , $x^2$ , $x^3$
4	Trigonometric	$\sin 4x$ , $\cos x$
5	Exponential	$e^{3x}$ , $e^{-x}$

When the integral reappears, solve for it.
Step	Working
First parts	$I = \displaystyle\int e^x\sin 2x\,dx = e^x\sin 2x - 2\int e^x\cos 2x\,dx$
Second parts	$\displaystyle\int e^x\cos 2x\,dx = e^x\cos 2x + 2I$
Combine	$I = e^x\sin 2x - 2e^x\cos 2x - 4I$
Solve	$I = \dfrac{e^x}{5}(\sin 2x - 2\cos 2x) + C$

Formulas

Integration by parts
$\int u\,dv = uv - \int v\,du$

Key terms

First function: The factor $u$ that is differentiated.
Second function: The factor $dv$ that is integrated.

2. The pattern e^x [f(x) + f'(x)]

Since $\dfrac{d}{dx}\left[e^x f(x)\right] = e^x\left[f(x) + f'(x)\right]$ , we have $\displaystyle\int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) + C$ . Rewrite the bracket if necessary, for example $\dfrac{x + 2}{(x + 3)^2} = \dfrac{1}{x + 3} - \dfrac{1}{(x + 3)^2}$ .

Spot $f$ and $f'$ side by side, then write $e^x f(x) + C$.
$f(x)$	$f'(x)$	$\int e^x[f(x) + f'(x)]\,dx$
$\dfrac1x$	$-\dfrac1{x^2}$	$\dfrac{e^x}{x} + C$
$\cos x$	$-\sin x$	$e^x\cos x + C$
$\tan x$	$\sec^2 x$	$e^x\tan x + C$
$\dfrac{1}{x + 3}$	$-\dfrac{1}{(x + 3)^2}$	$\dfrac{e^x}{x + 3} + C$

Formulas

Exponential pattern
$\int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) + C$

Key terms

Pattern: An exponential times a function plus its derivative.

3. Partial fractions

For a proper fraction, each linear factor $(ax + b)$ gives $\dfrac{A}{ax + b}$ ; a repeated factor $(ax + b)^2$ gives $\dfrac{A}{ax + b} + \dfrac{B}{(ax + b)^2}$ ; an irreducible quadratic gives $\dfrac{Ax + B}{x^2 + px + q}$ . An improper fraction is divided first. Find constants by the cover-up rule, then by comparing coefficients; check by recombining at a sample value.

One unknown constant for each degree of the denominator.
Factor in the denominator	Partial-fraction terms
Linear $(ax + b)$	$\dfrac{A}{ax + b}$
Repeated $(ax + b)^2$	$\dfrac{A}{ax + b} + \dfrac{B}{(ax + b)^2}$
Quadratic $x^2 + px + q$ (no real roots)	$\dfrac{Ax + B}{x^2 + px + q}$
Degree of numerator ≥ degree of denominator	Divide first, then split the remainder

Imagined example; constants recombined at sample points.
Step	Working
Form	$\dfrac{x + 11}{(x - 1)(x + 3)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 3}$
Cover-up	$x = 1$ : $A = \dfrac{12}{4} = 3$ ; $x = -3$ : $B = \dfrac{8}{-4} = -2$
Integrate	$3\log\|x - 1\| - 2\log\|x + 3\| + C$

A repeated factor gives a power term that integrates without a logarithm.
Step	Working
Form	$\dfrac{1}{(x + 1)(x - 2)^2} = \dfrac{A}{x + 1} + \dfrac{B}{x - 2} + \dfrac{C}{(x - 2)^2}$
Constants	$A = \dfrac19$ ( $x = -1$ ); $C = \dfrac13$ ( $x = 2$ ); $x^2$ terms: $A + B = 0$ , so $B = -\dfrac19$
Integrate	$\dfrac19\log\|x + 1\| - \dfrac19\log\|x - 2\| - \dfrac{1}{3(x - 2)} + C$

Formulas

Two linear factors
$\frac{px + q}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b}$

Key terms

Proper fraction: Degree of numerator less than degree of denominator.
Cover-up rule: For a simple linear factor, delete it and put its root into what remains.

4. Quadratic denominators and their roots

Complete the square and use: $\displaystyle\int \dfrac{dx}{x^2 + a^2} = \dfrac1a\tan^{-1}\dfrac{x}{a}$ ; $\displaystyle\int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a}\log\left|\dfrac{x - a}{x + a}\right|$ ; $\displaystyle\int \dfrac{dx}{a^2 - x^2} = \dfrac{1}{2a}\log\left|\dfrac{a + x}{a - x}\right|$ ; $\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\dfrac{x}{a}$ ; $\displaystyle\int \dfrac{dx}{\sqrt{x^2 \pm a^2}} = \log\left|x + \sqrt{x^2 \pm a^2}\right|$ (each $+\,C$ ). For $\dfrac{px + q}{ax^2 + bx + c}$ (or its root form), write $px + q = \lambda(2ax + b) + \mu$ .

Learn these five; every quadratic form reduces to one of them.
Integrand	Integral
$\dfrac{1}{x^2 + a^2}$	$\dfrac1a\tan^{-1}\dfrac{x}{a} + C$
$\dfrac{1}{x^2 - a^2}$	$\dfrac{1}{2a}\log\left\|\dfrac{x - a}{x + a}\right\| + C$
$\dfrac{1}{a^2 - x^2}$	$\dfrac{1}{2a}\log\left\|\dfrac{a + x}{a - x}\right\| + C$
$\dfrac{1}{\sqrt{a^2 - x^2}}$	$\sin^{-1}\dfrac{x}{a} + C$
$\dfrac{1}{\sqrt{x^2 \pm a^2}}$	$\log\left\|x + \sqrt{x^2 \pm a^2}\right\| + C$

Halve the coefficient of $x$, square it, add and subtract.
Quadratic	Completed square	Form reached
$x^2 + 6x + 13$	$(x + 3)^2 + 2^2$	$u^2 + a^2$
$x^2 - 4x + 13$	$(x - 2)^2 + 3^2$	$u^2 + a^2$
$8 + 2x - x^2$	$3^2 - (x - 1)^2$	$a^2 - u^2$
$2x^2 + 8x + 10$	$2\left[(x + 2)^2 + 1\right]$	take out the 2 first

Numerator = multiple of the derivative + constant.
Step	Working
Split	$3x - 1 = \dfrac32(2x + 2) - 4$
Log part	$\dfrac32\displaystyle\int \dfrac{2x + 2}{x^2 + 2x + 5}\,dx = \dfrac32\log(x^2 + 2x + 5)$
Constant part	$-4\displaystyle\int \dfrac{dx}{(x + 1)^2 + 4} = -2\tan^{-1}\dfrac{x + 1}{2}$
Answer	$\dfrac32\log(x^2 + 2x + 5) - 2\tan^{-1}\dfrac{x + 1}{2} + C$

Formulas

Linear numerator split
$px + q = \lambda\,\frac{d}{dx}(ax^2 + bx + c) + \mu$

Key terms

Completing the square: Writing $x^2 + bx + c$ as $\left(x + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right)$ .

5. Square roots of quadratics

$\displaystyle\int \sqrt{a^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$ ; $\displaystyle\int \sqrt{x^2 \pm a^2}\,dx = \dfrac{x}{2}\sqrt{x^2 \pm a^2} \pm \dfrac{a^2}{2}\log\left|x + \sqrt{x^2 \pm a^2}\right| + C$ . For $(px + q)\sqrt{ax^2 + bx + c}$ , split $px + q$ as above: the derivative part gives a power $\dfrac23(\ldots)^{3/2}$ , the constant part one of these three results.

Same first term each time; the second term changes.
Integrand	Integral
$\sqrt{a^2 - x^2}$	$\dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$
$\sqrt{x^2 + a^2}$	$\dfrac{x}{2}\sqrt{x^2 + a^2} + \dfrac{a^2}{2}\log\left\|x + \sqrt{x^2 + a^2}\right\| + C$
$\sqrt{x^2 - a^2}$	$\dfrac{x}{2}\sqrt{x^2 - a^2} - \dfrac{a^2}{2}\log\left\|x + \sqrt{x^2 - a^2}\right\| + C$

$y = \sqrt{9 - x^2}$

The curve is a semicircle; this is why $\sin^{-1}$ appears in its integral.

Formulas

Root of a sum of squares
$\int \sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log\left|x + \sqrt{x^2 + a^2}\right| + C$

Key terms

Root form: An integrand containing the square root of a quadratic.

6. Trigonometric forms and the special integrals

With $t = \tan\dfrac{x}{2}$ : $\sin x = \dfrac{2t}{1 + t^2}$ , $\cos x = \dfrac{1 - t^2}{1 + t^2}$ , $dx = \dfrac{2\,dt}{1 + t^2}$ — for $\dfrac{1}{a + b\cos x}$ , $\dfrac{1}{a + b\sin x}$ , $\dfrac{1}{a\cos x + b\sin x + c}$ . For $\dfrac{1}{a\cos^2 x + b\sin^2 x + c}$ divide by $\cos^2 x$ and put $t = \tan x$ . For $\dfrac{c\cos x + d\sin x}{a\cos x + b\sin x}$ , write the numerator as $A(\text{den}) + B(\text{den})'$ . Special integrals: divide $\dfrac{1 \pm x^2}{1 + x^4}$ by $x^2$ and put $t = x \mp \dfrac1x$ ; for $\sqrt{\tan x}$ or $\sqrt{\cot x}$ put $\tan x = t^2$ .

Every trigonometric form in the syllabus becomes a rational function of $t$.
With $t = \tan\dfrac{x}{2}$	Replace by
$\sin x$	$\dfrac{2t}{1 + t^2}$
$\cos x$	$\dfrac{1 - t^2}{1 + t^2}$
$dx$	$\dfrac{2\,dt}{1 + t^2}$
$a\cos^2 x + b\sin^2 x + c$ forms	divide by $\cos^2 x$ ; put $t = \tan x$

For $x > 0$; root tan and root cot on $\left(0, \tfrac{\pi}{2}\right)$.
Integrand	Divide by $x^2$ ; substitute	Result
$\dfrac{1 + x^2}{1 + x^4}$	$t = x - \dfrac1x$ ; $t^2 + 2$	$\dfrac{1}{\sqrt2}\tan^{-1}\dfrac{x^2 - 1}{\sqrt2\,x} + C$
$\dfrac{1 - x^2}{1 + x^4}$	$t = x + \dfrac1x$ ; $t^2 - 2$	$\dfrac{1}{2\sqrt2}\log\left\|\dfrac{x^2 + \sqrt2 x + 1}{x^2 - \sqrt2 x + 1}\right\| + C$
$\sqrt{\tan x}$ , $\sqrt{\cot x}$	$\tan x = t^2$ , then the two forms above	sum or difference of the two results

A scaled version of the special form; $x > 0$.
Step	Working
Divide by $x^2$	$\dfrac{1 + \frac{4}{x^2}}{x^2 + \frac{16}{x^2}}$
Substitute	$t = x - \dfrac4x$ : $dt = \left(1 + \dfrac{4}{x^2}\right)dx$ , $x^2 + \dfrac{16}{x^2} = t^2 + 8$
Integrate	$\displaystyle\int \dfrac{dt}{t^2 + 8} = \dfrac{1}{2\sqrt2}\tan^{-1}\dfrac{t}{2\sqrt2}$
Answer	$\dfrac{1}{2\sqrt2}\tan^{-1}\left(\dfrac{x^2 - 4}{2\sqrt2\,x}\right) + C$

Formulas

Special integral
$\int \frac{1 + x^2}{1 + x^4}\,dx = \frac{1}{\sqrt2}\tan^{-1}\frac{x^2 - 1}{\sqrt2\,x} + C$

Key terms

Half-angle substitution: $t = \tan\frac{x}{2}$ , which makes $\sin x$ , $\cos x$ and $dx$ rational in $t$ .

Exam tips

How ISC marks are awarded

By parts (2–3 marks): choice of $u$ and $dv$ , the formula applied correctly, the final answer.
Partial fractions (2–5 marks): the correct form, the constants, the integration.
Quadratic and root forms (2–3 marks): completed square, standard result, answer.
Trigonometric forms (3–5 marks): substitution, simplified integral, answer in $x$ .
Section D (5 marks): questions come in parts such as 2 + 3; label each part.

Common mistakes

Choosing $u$ against the order (for example differentiating $e^x$ and integrating $\log x$ ).
Too few constants for a repeated factor, or a constant numerator over a quadratic factor.
Forgetting to divide an improper fraction first.
Losing the factor taken out of $ax^2 + bx + c$ when completing the square.
Mixing up the factors $\dfrac1a$ and $\dfrac{1}{2a}$ , or the sign of the log term in $\int\sqrt{x^2 - a^2}\,dx$ .

1. Integration by parts

Formulas

Key terms

2. The pattern e^x [f(x) + f'(x)]

Formulas

Key terms

3. Partial fractions

Formulas

Key terms

4. Quadratic denominators and their roots

Formulas

Key terms

5. Square roots of quadratics

Formulas

Key terms

6. Trigonometric forms and the special integrals

Formulas

Key terms

Exam tips

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