Question

Integrate:

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

Answer 1

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

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

$\Rightarrow \int \frac{1}{x(1-2x)}dx-\int \frac{x}{1-2x}dx$

Let $M=\int \frac{1}{x(1-2x)}$ & $N=\int \frac{x}{1-2x}dx$

$\Rightarrow$ using partial fraction in M we get

$1=A(1-2x)+Bx$

$1=x(-2A+B)+A$

$[A=1]$ & $-2A+B=0$

$[B=2A=2]$

Hence

$M=\int \frac{1}{x}+\frac{2}{1-2x}dx=logx+2log(1-2x)$ & $N=\int \frac{x}{1-2x}dx$

Let $1-2x=v$ & $x=(1-v)/2$

$-2dx=dv$

$N=\frac{-1}{4}\int 1-v.dv=\frac{-1}{4}[v-\frac{v^2}{2}]$

$=\frac{-1}{4}[1-2x-\frac{(1-2x{)}^2}{2}]$

$=-\frac{4x^2+1-4x+4x-2}{8}$

$=\frac{4x^2-1}{8}$

Hence

$P=M+N=log\frac{x}{(1-2x{)}^2}+\frac{4x^2-1}{8}+c$