∫1(x^2+a^2)^32 Substitute x=atan u , u=tan^ 1(xa) , #160; dx=a^2udu =∫a^2u(a^2tan^2u+a^2)^32du a^2tan^2u=a^2^2u =1a^2∫1 ud ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Integration by Partial Fractions

Solve ∫1√a2...

Question

Solve
$\int \frac{1}{\sqrt{(a^{2} + x^{2})^{3}}}$

Open in App

Solution

$\int \frac{1}{(x^{2} + a^{2})^{\frac{3}{2}}}$

Substitute $x = a tan u$ , $u = {tan}^{- 1} (\frac{x}{a})$ , $d x = a {sec}^{2} u d u$

$= \int \frac{a {sec}^{2} u}{(a^{2} {tan}^{2} u + a^{2})^{\frac{3}{2}}} d u$

$a^{2} {tan}^{2} u = a^{2} {sec}^{2} u$

$= \frac{1}{a^{2}} \int \frac{1}{sec u} d u$

$= \frac{1}{a^{2}} \int cos u d u$

$= \frac{sin u}{a^{2}}$

$= \frac{sin ({tan}^{-} 1 (\frac{x}{a}))}{a^{2}}$

$= \frac{x}{a^{3} \sqrt{\frac{x^{2}}{a^{2}} + 1}}$

$= \frac{x}{a^{3} \sqrt{\frac{x^{2} + a^{2}}{a^{2}}}}$

$= \frac{a x}{a^{3} \sqrt{x^{2} + a^{2}}}$

$= \frac{x}{a^{2} \sqrt{x^{2} + a^{2}}}$

Suggest Corrections