Evaluate the following integrals:

Let $I=\int \frac{1}{\sqrt{5-4 x-2 x^{2}}} d x$

$=\int \frac{1}{\sqrt{-2\left[x^{2}+2 x-\frac{5}{2}\right]}} d x$

$=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left[x^{2}+2 x+(1)^{2}-(1)^{2}-\frac{5}{2}\right]}} d x$

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

$=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\frac{7}{2}-(x+1)^{2}}} d x$

Let $(x+1)=t$

Differentiating both sides, we get,

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

So, $I=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\sqrt{\left(\frac{7}{2}\right)}\right)^{2}-t^{2}}} d t$

$=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{t}{\sqrt{\frac{7}{2}}}\right)+c$

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

$I=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\sqrt{\frac{2}{7}} \times(x+1)\right)+c$