Evaluating an integral: $\int \frac{1}{(x+\sqrt{x+x^2})^2} dx$

Question

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

I don't know how to approach this integral. I tried a few substitutions, but none of them got me to a desirable point.

Answer 1

We have if $x>0 $ $$\int\frac{1}{\left(x+\sqrt{x+x^{2}}\right)^{2}}dx=\int\frac{1}{x^{2}\left(1+\sqrt{1+\frac{1}{x}}\right)^{2}}dx $$ and now put $u=1+1/x $, $du=-dx/x^{2} $ and so we get $$-\int\frac{1}{\left(1+\sqrt{u}\right)^{2}}du. $$ Now put $s=\sqrt{u} $, $ds=1/\left(2\sqrt{u}\right)du $ and get $$-2\int\frac{s}{\left(s+1\right)^{2}}ds $$ which can be split using partial fractions $$-2\int\frac{1}{s+1}ds+2\int\frac{1}{\left(s+1\right)^{2}}ds. $$ I think you can get it from here. If $x<-1 $ you have $$-\int\frac{1}{\left(1-\sqrt{u}\right)^{2}}du $$ and you can use the same passage of the first part using $s=-\sqrt{u} $.