Evaluate: $\int \frac{2}{(1-x)(1+x^2)}dx$

Question

Given $\int \frac{2}{(1-x)(1+x^2)}dx$

The most obvious approach is to use Partial fractions

Let $\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^2}$

$\Rightarrow \frac{2}{(1-x)(1+x^2)}=\frac{A+Ax^2+Bx-Bx^2+C-Cx}{(1-x)(1+x^2)}$

We get $A=1, B=1, C=1$

The integral now becomes

$\int[\frac{1}{1-x}+\frac{x+1}{1+x^2}]dx$

$\Rightarrow \int[\frac{1}{1-x}+\frac{x}{1+x^2}+\frac{1}{1+x^2}]dx$

Now, we can simply integrate term by term

$\Rightarrow -\log|1-x|+\frac{\log(1+x^2)}{2}+\tan^{-1}x+c $

$\Rightarrow \log |\frac{\sqrt{1+x^2}}{1-x}|+\tan^{-1}x+c $

Please review my solution and if you have any other way of integrating then please share your solution.

Answer 1

Here is another way to integrate \begin{align} &\int \frac{2}{(1-x)(1+x^2)}dx\\ =& \int \frac{2(1+x)}{1-x^4}dx= \int \frac{2x}{1-x^4}+ \frac{2}{1-x^4} \ dx\\ =& \int \frac{1}{1-x^4} \ d(x^2)+\int \frac{1}{1-x^2}+\frac{1}{1+x^2}\ dx\\ = &\ \frac12\ln|\frac{1+x^2}{1-x^2}|+ \frac12\ln|\frac{1+x}{1-x}|+\tan^{-1}x+C \end{align}

Answer 2

$$\Im \left(\frac{1}{x-i}\right)=\frac{1}{1+x^2}$$ $$\int \frac{2}{(1-x)(1+x^2)}dx=\Im\int\frac{2}{(1-x)(x-i)}dx$$$$=\Im\int\frac{(1+i)((1-x)+(x-i))}{(1-x)(x-i)}dx=\Im\left[\int\frac{1+i}{x-i}dx+\int\frac{1+i}{1-x}dx\right]$$ $$=\Im\left[(1+i)\ln(x-i)-(1+i)\ln(1-x)\right]+c$$$$=\frac{1}{2}\ln(x^2+1)+\arctan x-\ln(1-x)+C$$