integrate

or use

eulers substitution which u only taught me !!!

Let
I_n=\int_0^\infty \dfrac{\mathrm dx}{(x+\sqrt{1+x^2})^n}
Integration by parts yield
I_n=\left|\dfrac{x}{(x+\sqrt{1+x^2})^n}\right|_0^\infty+n\int_0^\infty\dfrac{ x \ \mathrm dx}{(x+\sqrt{1+x^2})^n\sqrt{1+x^2}}
The first term yields 0 and so we get
I_n=n\int_0^\infty\dfrac{ x +\sqrt{1+x^2}-\sqrt{1+x^2} }{(x+\sqrt{1+x^2})^n\sqrt{1+x^2}}\ \mathrm dx
which further gives
I_n=n\int_0^\infty\dfrac{ x +\sqrt{1+x^2} }{(x+\sqrt{1+x^2})^n\sqrt{1+x^2}}\ \mathrm dx -nI_n
from where we get
(n+1)I_n=n\int_0^\infty\dfrac{ x +\sqrt{1+x^2} }{(x+\sqrt{1+x^2})^n\sqrt{1+x^2}}\ \mathrm dx
The integral could be done easily by the substitution
z = x+√1+x² which gives
dz = x+√1+x²√1+x² dx
So we get
(n+1)I_n=n\int_1^\infty \dfrac{ \mathrm dz}{z^n}=\dfrac{n}{n-1}
And so ultimately
\boxed{I_n=\dfrac{n}{n^2-1}}