Integration - indefinite integral of $\frac{1}{x \sqrt{x^2+4x}}$

Question

$$ \int \frac{1}{x \sqrt{x^2+4x}} dx $$

I have tried different substitutions writing the denominator as $x \sqrt{(x+2)^2-4}$ but can't make it work.

Answer 1

Assuming, $x>0$, write the integrand as $$ \frac{1}{\sqrt{1+4/x}}\frac{1}{x^2}, $$ and let $t=1/x$. You get $$ -\int\frac{1}{\sqrt{1+4t}}\,dt, $$ which I'm sure you can handle. Also, once you are done, try to think about the case of negative $x$.