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.

Sub $x \mapsto 1/u$...
$$\int_1^{\infty} \frac{dx}{x^3 \sqrt{x^2+x}} = \int_0^1 du \frac{u^2}{\sqrt{1+u}} = \int_1^2 dv \, v^{-1/2} (v-1)^2 = \int_1^2 dv \, \left (v^{3/2} - 2 v^{1/2} + v^{-1/2} \right )$$
which is
$$\frac25 \left (2^{5/2} - 1 \right ) -2 \frac23 \left (2^{3/2} - 1 \right ) + 2 \left (2^{1/2}-1 \right ) = \frac{14}{15} \sqrt{2} - \frac{16}{15}$$
