Integral proof $I_n=\int\frac{x^n}{\sqrt{x^2+a}} \, dx$ If $$I_n=\int\frac{x^n}{\sqrt{x^2+a}}\,dx$$ prove that $$I_n=\frac{1}{n}x^{n-1}\sqrt{x^2+a}-\frac{n-1}{n}aI_{n-2}$$
I tried
 $$I_n=\int\frac{x\times x^{n-1}}{\sqrt{x^2+a}} \,dx =\int(\sqrt{x^2+a})' \times x^{n-1} \,dx \\ =x^{n-1}\sqrt{x^2+a} -(n-1)\int x^{n-2}\sqrt{x^2+a} \, dx$$
I cant any further. Any hint?

Note that
\begin{align}
\int\frac{x^n}{\sqrt{x^2+a}}\,dx
= &\int \frac{x^{n-1}}{n(x^2+a)^{\frac{n-1}2}}\ d\left[(x^2+a)^{\frac n2}\right]
\end{align}
Then, integrate by parts to immediately establish
$$I_n= \frac{x^{n-1}}{n}\sqrt{x^2+a}-\frac{(n-1)a}{n}I_{n-2}
$$
