Integrate $\frac1{1 + ax^2}$ using partial fractions decomposition

Question

I'm trying to solve the following integral:

Integral of: $\int\frac{1}{1+ax^2}dx$

Where $a$ is some positive constant.

We can't of course use basic U Substitution as the derivative of $1 + ax^2$ is $2ax$, which isn't found elsewhere in the expression. I tried to use fractional decomposition to integrate this but ran into the issue outlined below.

I'd like to divide this into two related questions:

1. My first approach was to try and use fractional decomposition to integrate this. However, I wasn't able to factor the expression $1 + ax^2$ to a product of two linear expressions. Can this be done?

2. If we can't do fractional decomposition, how can this integral be solved?

Answer 1

This is an immediate integral of a composition of function. Since $a>0$,

$$ \int \frac{dx}{1+ax^2} = \frac{1}{\sqrt{a}}\int\frac{\sqrt{a}}{1+(\sqrt{a}x)^2}dx = \frac{1}{\sqrt{a}}\arctan(\sqrt{a}x)+c $$

Answer 2

(1) You can factor the denominator into linear factors and integrate from there, but doing so necessarily involves a detour into complex numbers. The denominator factors over $\Bbb C$, in our case as $$(1 + i \sqrt a x) (1 + i \sqrt a x) .$$ Applying the Method of Partial Fractions gives $$\int \frac{dx}{1 + a x^2} = \frac{i}{2} \int \left(\frac{1}{\sqrt a x + i} - \frac{1}{\sqrt a x - i}\right) \,dx = \frac{1}{\sqrt a} \cdot \frac{i}{2} \log \frac{\sqrt{a} x + i}{\sqrt{a} x - i} + C .$$ But using the identity $$\frac{i}{2} \log \frac{z + i}{z - i} = \arctan z ,$$ yields the more familiar form $$\boxed{\int \frac{dx}{1 + a x^2} = \frac{1}{\sqrt a} \arctan \sqrt a x + C} .$$

(2) Taking $u = \sqrt a x$, so that the denominator becomes $1 + u^2$, transforms the integral to $$\frac{1}{\sqrt a} \int \frac{du}{1 + u^2} = \frac{1}{\sqrt a} \arctan u + C = \frac{1}{\sqrt a} \arctan \sqrt a x + C .$$