Evaluate the following integrals:

let $I=\int \frac{1}{2 x^{2}-x-1} d x$

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

$=\frac{1}{2} \int \frac{1}{x^{2}+2 x \times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}-\frac{1}{2}} d x$

$=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\frac{9}{16}} d x$

Let $\left(x-\frac{1}{4}\right)=t \ldots \ldots$ (i)

$\Rightarrow \mathrm{d} x=\mathrm{dt}$

SO,

$I=\frac{1}{2} \int \frac{1}{t^{2}-\left(\frac{3}{4}\right)^{2}} d t$

$I=\frac{1}{2} \times \frac{1}{2 \times \frac{3}{4}} \log \left|\frac{t-\frac{3}{4}}{t+\frac{3}{4}}\right|+c$

[since, $\left.\int \frac{1}{x^{2}-(a)^{2}} d x=\frac{1}{2 \times a} \log \left|\frac{x-a}{x+a}\right|+c\right]$

$I=\frac{1}{3} \log \left|\frac{x-\frac{1}{4}-\frac{3}{4}}{x-\frac{1}{4}+\frac{3}{4}}\right|+c$ [using $\left.(i)\right]$

$I=\frac{1}{3} \log \left|\frac{x-1}{2 x+1}\right|+c$