How to calculate this definite integral? How can I calculate this 
$$\int_1^{+\infty} \frac {dx}{x^3\sqrt{x^2+x}}$$
I have no idea what to do with it, integration by parts or by substitution doesn't work for me.

We have, as $x \to 0^+$,
$$
 \frac 1{x^3\sqrt{x^2+x}}\sim  \frac 1{x^{7/2}}
$$
and since $7/2>1$, the integral $\displaystyle \int_0^{+\infty} \frac {dx}{x^3\sqrt{x^2+x}}$ is not convergent.

Since the OP has changed the initial integral, a route for the new one is to make the change of variable $u=\dfrac1x$.
