Integral $ \int_a^\infty \frac{dr}{r^5 \sqrt{r-a}}$ 
$$ \int_a^\infty \frac{dr}{r^5 \sqrt{r-a}}$$

By taking $r-a=t^2$ we are getting 
 $$ \int_0^\infty \frac{2dt}{(a+t^2)^5}$$
Now I am stuck after this. Thanks.

Let $t = \sqrt{a}\tan(\theta), dt = \sqrt{a}\sec^2(\theta)$. Then your integral becomes
$$2\int \frac{\sqrt{a}\sec^2(\theta)}{(a+a\tan^2(\theta))^5} d \theta = \frac{2\sqrt a}{a^5} \int \cos^8(\theta)\, d\theta$$
$$ = \frac{2}{a^{9/2}} \int \left( \frac{1+\cos(2\theta)}{2}\right)^4 d\theta$$$$ = \frac{1}{8a^{9/2}}\int 1+4\cos(2\theta)+6\cos^2(2\theta)+4\cos^3(2\theta)+\cos^4(2\theta) d\theta$$
