Integral Resulting in Arctangent

Theorem

$\displaystyle \int \frac 1 {x^2 + a^2} \ \mathrm dx = \frac 1 a \arctan{\frac x a} + C$

where $a$ is a non-zero constant.

Proof

Substitute:

$a \tan \theta = x$

for $\theta \in \left({-\dfrac \pi 2 .. \dfrac \pi 2}\right)$.

From Shape of Tangent Function, this substitution is valid for all real $x$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a \tan \theta\)	\(=\)	\(\displaystyle x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a \sec^2 \theta \frac{\mathrm d \theta}{\mathrm dx}\)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	differentiate WRT $x$, Derivative of Tangent Function, Chain Rule
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \int \frac 1 {x^2 + a^2} \ \mathrm d x\)	\(=\)	\(\displaystyle \int \frac {a \ \sec^2 \theta} {a^2 \tan^2 \theta + a^2} \frac {\mathrm d \theta}{\mathrm d x} \mathrm d x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Integration by Substitution
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac a {a^2} \int \frac {\sec^2 \theta} {\tan^2 \theta + 1} \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Algebraic rearrangement
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 a \int \frac {\sec^2 \theta} {\sec^2 \theta} \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	corollary to sum of squares of sine and cosine
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 a \int \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 a \theta + C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Integration of a Constant

As $\theta$ was stipulated to be in the open interval $\left({-\dfrac \pi 2 .. \dfrac \pi 2}\right)$:

$\tan \theta = \dfrac x a \iff \theta = \arctan \dfrac x a$

The answer in terms of $x$, then, is:

$\displaystyle \int \frac 1 {x^2 + a^2} \ \mathrm dx = \frac 1 a \arctan \frac x a + C$

$\blacksquare$

Also see

• Derivative of Arctangent Function

Sources

• For a video presentation of the contents of this page, visit the Khan Academy.