Question

Integrate the rational functions.
$\int \frac{2}{(1-x)(1+x^2)}dx$

Answer 1

Let $\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^2}$
$\therefore \frac{2}{(1-x)(1+x^2)}=\frac{A(1+x^2)+(Bx+C)(1-x)}{(1-x)(1+x^2)}$
$\Rightarrow 2=A+A{x}^2+Bx+C-B{x}^2-Cx\Rightarrow 2=x^2(A-B)+x(B-C)+(A+C)$
On comparing the coefficients of $x^2$, x and constant term on both sides, we get
$A-B=0\Rightarrow A=B.......\left(i\right)B-C=0\Rightarrow B-C........\left(ii\right)$
and $A+C=2........\left(iii\right)$
From Eqs. (i) and (ii), we get A =C put this value in Eq. (iii), we get
$2A=2\Rightarrow A=1$
Put the value of A in Eqs. (i) and (iii), we get B =1 and C =1
$\therefore \int \frac{2}{(1-x)(1+x^2)}dx=\int \{\frac{1}{1-x}+\frac{x+1}{x^2+1}\}dx\int \frac{1}{1-x}dx+\frac{1}{2}\int \frac{2x}{x^2+1}dx+\int \frac{1}{x^2+1}dx=-log|1-x|+\frac{1}{2}log(1+x^2)+ta{n}^{-1}x+C[\text{Let}{x}^2+1=t\Rightarrow 2xdx=dt\therefore \int \frac{2x}{x^2+1}dx=\int \frac{1}{t}dt=logt]$