Question

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

Answer 1

Let $I=\frac{2}{(1-x)(1+x^2)}dx$

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

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

$\Rightarrow 2=A(1+x^2)+(Bx+C)(1-x)$

$\Rightarrow 2=A+A{x}^2+Bx-B{x}^2+C-Cx$

$\Rightarrow 2=(A+C)+(A-B)x^2+(B-C)x$

Equating coefficients both sides, we get

$A+C=2,A-B=0,B-C=0$

$\Rightarrow A=B,B=C$

$\Rightarrow A=B=C,A+A=2\Rightarrow A=1$

therefore $A=B=C=1$

So, $\frac{2}{(1-x)(1+x^2)}=\frac{1}{1-x}+\frac{x+1}{1+x^2}$

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

$=-log|1-x|+\int \frac{x}{1+x^2}dx+\int \frac{1}{1+x^2}dx$

$=-log|1-x|+\frac{1}{2}log|1+x^2|+ta{n}^{-1}x+C$