Integration of Particular Functions

Two standard forms

A handful of rational forms show up often enough in calculus that it's worth knowing their antiderivatives directly, rather than re-deriving them via partial fractions every time:

\int \dfrac{dx}{x^2-a^2} = \dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right| + C

\int \dfrac{dx}{a^2-x^2} = \dfrac{1}{2a}\ln\left|\dfrac{a+x}{a-x}\right| + C

The full set of standard forms

Seven more forms complete the set. All nine appear in the NCERT Class 12 textbook, and each one is worth knowing by heart.

The nine standard integrals of particular functions

Integral	Result
$\displaystyle\int \dfrac{dx}{x^2 - a^2}$	$\dfrac{1}{2a}\ln\left\|\dfrac{x - a}{x + a}\right\| + C$
$\displaystyle\int \dfrac{dx}{a^2 - x^2}$	$\dfrac{1}{2a}\ln\left\|\dfrac{a + x}{a - x}\right\| + C$
$\displaystyle\int \dfrac{dx}{x^2 + a^2}$	$\dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$
$\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}}$	$\ln\left\|x + \sqrt{x^2 - a^2}\right\| + C$
$\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}}$	$\sin^{-1}\dfrac{x}{a} + C$
$\displaystyle\int \dfrac{dx}{\sqrt{x^2 + a^2}}$	$\ln\left\|x + \sqrt{x^2 + a^2}\right\| + C$
$\displaystyle\int \sqrt{a^2 - x^2}\,dx$	$\dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$
$\displaystyle\int \sqrt{x^2 + a^2}\,dx$	$\dfrac{x}{2}\sqrt{x^2 + a^2} + \dfrac{a^2}{2}\ln\left\|x + \sqrt{x^2 + a^2}\right\| + C$
$\displaystyle\int \sqrt{x^2 - a^2}\,dx$	$\dfrac{x}{2}\sqrt{x^2 - a^2} - \dfrac{a^2}{2}\ln\left\|x + \sqrt{x^2 - a^2}\right\| + C$

One example of each form

Each standard form applied once, with the value of a read off

Integral	Form used	Result
$\displaystyle\int \dfrac{dx}{x^2 - 16}$	$x^2 - a^2$ , $a = 4$	$\dfrac{1}{8}\ln\left\|\dfrac{x - 4}{x + 4}\right\| + C$
$\displaystyle\int \dfrac{dx}{9 - x^2}$	$a^2 - x^2$ , $a = 3$	$\dfrac{1}{6}\ln\left\|\dfrac{3 + x}{3 - x}\right\| + C$
$\displaystyle\int \dfrac{dx}{x^2 + 25}$	$x^2 + a^2$ , $a = 5$	$\dfrac{1}{5}\tan^{-1}\dfrac{x}{5} + C$
$\displaystyle\int \dfrac{dx}{\sqrt{x^2 - 4}}$	$\sqrt{x^2 - a^2}$ , $a = 2$	$\ln\left\|x + \sqrt{x^2 - 4}\right\| + C$
$\displaystyle\int \dfrac{dx}{\sqrt{1 - 4x^2}}$	$\sqrt{a^2 - x^2}$ after taking out the 2	$\dfrac{1}{2}\sin^{-1}(2x) + C$
$\displaystyle\int \dfrac{dx}{\sqrt{x^2 + 9}}$	$\sqrt{x^2 + a^2}$ , $a = 3$	$\ln\left\|x + \sqrt{x^2 + 9}\right\| + C$
$\displaystyle\int \sqrt{16 - x^2}\,dx$	$\sqrt{a^2 - x^2}$ , $a = 4$	$\dfrac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\dfrac{x}{4} + C$
$\displaystyle\int \sqrt{x^2 + 4}\,dx$	$\sqrt{x^2 + a^2}$ , $a = 2$	$\dfrac{x}{2}\sqrt{x^2 + 4} + 2\ln\left\|x + \sqrt{x^2 + 4}\right\| + C$
$\displaystyle\int \sqrt{x^2 - 9}\,dx$	$\sqrt{x^2 - a^2}$ , $a = 3$	$\dfrac{x}{2}\sqrt{x^2 - 9} - \dfrac{9}{2}\ln\left\|x + \sqrt{x^2 - 9}\right\| + C$

For $\sqrt{1 - 4x^2}$ , write it as $2\sqrt{\tfrac{1}{4} - x^2}$ first. Then $a = \dfrac{1}{2}$ , and the 2 taken out becomes the $\dfrac{1}{2}$ in front of the answer.

Worked example

Find $\displaystyle\int \dfrac{x^2}{1-x^6}\,dx$ .

Solution: Substitute $u = x^3$ , so $du = 3x^2\,dx$ :

\int \dfrac{x^2}{1-x^6}\,dx = \dfrac{1}{3}\int \dfrac{du}{1-u^2} = \dfrac{1}{3}\cdot\dfrac{1}{2}\ln\left|\dfrac{1+u}{1-u}\right| + C = \dfrac{1}{6}\ln\left|\dfrac{1+x^3}{1-x^3}\right| + C

Worked example: completing the square

Find $\displaystyle\int \dfrac{dx}{x^2 + 6x + 13}$ .

Solution: The quadratic doesn't factor, so complete the square: $x^2 + 6x + 13 = (x + 3)^2 + 4$ . With $u = x + 3$ and $a = 2$ , this is the $x^2 + a^2$ form:

\int \dfrac{dx}{(x+3)^2 + 2^2} = \dfrac{1}{2}\tan^{-1}\dfrac{x + 3}{2} + C

Common mistakes

Ignoring a coefficient on $x^2$ . Take it out first, as with $\sqrt{1 - 4x^2}$ above, or the value of $a$ and the factor in front will both be wrong.
Swapping $x^2 - a^2$ and $a^2 - x^2$ . They give the same logarithm upside down, so the sign of the answer flips.
Not completing the square. A quadratic with an $x$ term, like $x^2 + 6x + 13$ , has to be rewritten as $(x + h)^2 \pm k^2$ before any form applies.