Question

$\int \frac{1}{{(2ax+x^2)}^{3/2}}dx=-\frac{1}{a^2}\frac{x+a}{\sqrt{x^2+2ax}}$.

If this is true enter 1, else enter 0.

Answer 1

Let $I=\int \frac{1}{{(2ax+x^2)}^{3/2}}dx=\int \frac{1}{{[{(x+a)}^2-a^2]}^{3/2}}dx$
Put $x+a=a\sec \theta \Rightarrow dx=\sec \theta \tan \theta d\theta$
Therefore
$I=\int \frac{a\sec \theta \tan \theta }{a^3{\left({\tan }^2\theta \right)}^{3/2}}d\theta =\frac{1}{a^2}\int \frac{1}{\cos \theta }\cdot \frac{{\cos }^2\theta }{{\sin }^2\theta }d\theta$
$=\frac{1}{a^2}\int \frac{\cos \theta }{{\sin }^2\theta }d\theta =\frac{1}{a^2}\cdot \frac{-1}{\sin \theta }=-\frac{1}{a^2}\cdot \frac{x+a}{\sqrt{x^2+2ax}}$