Question

$\int \frac{1}{(x+1)\sqrt{x+2}}dx=$

Answer 1

Let, $I=$$\int \frac{1}{(x+1)\sqrt{x+2}}dx$

Put $x+2=z^2$ then $dx=2zdz$.

Using these in the above integration we get,

$I=$$\int \frac{2zdz}{(z^2-1)z}dx$

or, $I=$$2\int \frac{dz}{(z^2-1)}dx$

or, $I=$$2.\frac{1}{2.1}\log \left|\frac{z-1}{z+1}\right|+c$ [ Where $c$ integrating constant]

or, $I=$$\log \left|\frac{\sqrt{x+2}-1}{\sqrt{x+2}+1}\right|+c$