# Evaluate the following integrals:

Question:

Evaluate the following integrals:

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

Solution:

Given $I=\int \frac{1-x^{2}}{(1-2 x) x} d x$

Rewriting, we get $\int \frac{x^{2}-1}{x(2 x-1)} d x$

Expressing the integral $\int \frac{\mathrm{P}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}=\int \mathrm{Q}(\mathrm{x}) \mathrm{dx}+\int \frac{\mathrm{R}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}$

$\Rightarrow \int \frac{x^{2}-1}{x(2 x-1)} d x=\int\left(\frac{x-2}{2 x(2 x-1)}+\frac{1}{2}\right) d x$

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

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

By partial fraction decomposition,

$\Rightarrow \frac{\mathrm{x}-2}{\mathrm{x}(2 \mathrm{x}-1)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{2 \mathrm{x}-1}$

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

$\Rightarrow x-2=2 A x-A+B x$

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

$\therefore A=2$ and $2 A+B=1$

$\therefore B=1-4=-3$

Thus, $\Rightarrow \frac{\mathrm{x}-2}{\mathrm{x}(2 \mathrm{x}-1)}=\frac{2}{\mathrm{x}}-\frac{3}{2 \mathrm{x}-1}$

$\Rightarrow \int\left(\frac{2}{x}-\frac{3}{2 x-1}\right) d x$

$\Rightarrow 2 \int \frac{1}{x} d x-3 \int \frac{1}{2 x-1} d x$

Consider $\int \frac{1}{\mathrm{x}} \mathrm{dx}$

We know that $\int \frac{1}{x} d x=\log |x|+c$

$\Rightarrow \int \frac{1}{x} d x=\log |x|$

And consider $\int \frac{1}{2 x-1} d x$

Let $u=2 x-1 \rightarrow d x=1 / 2 d u$

$\Rightarrow \int \frac{1}{2 x-1} d x=\frac{1}{2} \int \frac{1}{u} d u$

We know that $\int \frac{1}{x} d x=\log |x|+c$

$\Rightarrow \frac{1}{2} \int \frac{1}{\mathrm{u}} \mathrm{du}=\frac{\log |\mathrm{u}|}{2}=\frac{\log |2 \mathrm{x}-1|}{2}$

Then,

$\Rightarrow \int \frac{\mathrm{x}-2}{\mathrm{x}(2 \mathrm{x}-1)} \mathrm{dx}=2 \int \frac{1}{\mathrm{x}} \mathrm{dx}-3 \int \frac{1}{2 \mathrm{x}-1} \mathrm{dx}$

$=2(\log |x|)-3\left(\frac{\log |2 x-1|}{2}\right)$

Then,

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

$=\frac{1}{2}\left(2(\log |x|)-3\left(\frac{\log |2 x-1|}{2}\right)\right)+\frac{1}{2} \int 1 d x$

We know that $\int 1 \mathrm{dx}=\mathrm{x}+\mathrm{c}$

$\Rightarrow \log |x|-\frac{3 \log |2 x-1|}{4}+\frac{x}{2}+c$

$\therefore I=\int \frac{1-x^{2}}{(1-2 x) x} d x=-\frac{3 \log |2 x-1|}{4}+\log |x|+\frac{x}{2}+c$