Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{\sqrt{7-3 x-2 x^{2}}} d x$

Solution:

let $I=\int \frac{1}{\sqrt{7-3 x-2 x^{2}}} d x$

$=\int \frac{1}{\sqrt{-2\left[x^{2}+\frac{3}{2} x-\frac{7}{2}\right]}} d x$

$=\frac{1}{\sqrt{2}} \int \frac{1}{\left.\sqrt{-\left[x^{2}+2 x\left(\frac{3}{4}\right)+\left(\frac{3}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}-\frac{7}{2}\right.}\right]} d x$

$=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left[\left(x-\frac{3}{4}\right)^{2}-\frac{65}{16}\right]}} d x$

$=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{\sqrt{65}}{4}\right)^{2}-\left(x+\frac{3}{4}\right)^{2}}} d x$

let $\left(x+\frac{3}{4}\right)=t$

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

$I=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{\sqrt{65}}{4}\right)^{2}-(t)^{2}}} d t$

$=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{\mathrm{t}}{\frac{\sqrt{65}}{4}}\right)+\mathrm{c}$

$I=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{4\left(x+\frac{3}{4}\right)}{\sqrt{65}}\right)+c$

$I=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{4 x+3}{\sqrt{65}}\right)+c$