Evaluate the following integrals:

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

$\Rightarrow \int \frac{1}{(\sin x-2 \cos x)(2 \sin x+\cos x)} d x$

$=\int \frac{1}{2 \sin ^{2} x+\sin x \cos x-4 \sin x \cos x-2 \cos ^{2} x} d x$

Dividing the numerator and denominator by $\cos ^{2} x$,

$\Rightarrow \int \frac{1}{2 \sin ^{2} x+\sin x \cos x-4 \sin x \cos x-2 \cos ^{2} x} d x$

$=\int \frac{\sec ^{2} x}{2 \tan ^{2} x-3 \tan x-2} d x$

Putting $\tan x=t$ so that $\sec ^{2} x d x=d t$.

$\Rightarrow \int \frac{\sec ^{2} x}{2 \tan ^{2} x-3 \tan x-2} d x=\int \frac{d t}{2 t^{2}-3 t-2}$

$=\frac{1}{2} \int \frac{1}{\mathrm{t}^{2}-\frac{3}{2}-1} \mathrm{dt}$

$=\frac{1}{2} \int \frac{1}{\left(\mathrm{t}-\frac{3}{4}\right)^{2}-\left(\frac{5}{4}\right)^{2}} \mathrm{dt}$

We know that $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c$

$\Rightarrow \frac{1}{2} \int \frac{1}{\left(\mathrm{t}-\frac{3}{4}\right)^{2}-\left(\frac{5}{4}\right)^{2}} \mathrm{dt}=\frac{1}{2} \times \frac{1}{2\left(\frac{5}{4}\right)} \log \left|\frac{\mathrm{t}-\frac{3}{4}-\frac{5}{4}}{\mathrm{t}-\frac{3}{4}+\frac{5}{4}}\right|+\mathrm{c}$

$=\frac{1}{5} \log \left|\frac{t-2}{t+\frac{1}{2}}\right|+c$

$=\frac{1}{5} \log \left|\frac{2 \tan x-4}{2 \tan x+1}\right|+c$

$\therefore \mathrm{I}=\int \frac{1}{(\sin \mathrm{x}-2 \cos \mathrm{x})(2 \sin \mathrm{x}+\cos \mathrm{x})} \mathrm{dx}=\frac{1}{5} \log \left|\frac{2 \tan \mathrm{x}-4}{2 \tan \mathrm{x}+1}\right|+\mathrm{c}$