Evaluate the following integrals:
$\int \frac{\left(1-x^{2}\right)}{x(1-2 x)} d x$
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 $\int 1 \mathrm{dx}=\mathrm{x}+\mathrm{c}$
$\Rightarrow \log |x|-\frac{3 \log |2 x-1|}{4}+\frac{x}{2}+c$
$\therefore I=\int \frac{1-x^{2}}{(1-2 x) x} d x=-\frac{3 \log |2 x-1|}{4}+\log |x|+\frac{x}{2}+c$
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