Evaluate the indefinite integral $\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}$ without trigonometric substitution. In order to find
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}
$$
we set $t=\arctan x$. Then $x=\tan t$ and $dt=\frac{dx}{x^2+1}$, so
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}=\int\frac{dt}{\sqrt{\frac{1}{\cos^2t}}}=\int\cos tdt\\
=\sin t+C=\sin(\arctan x)+C
$$
Now, since
$$
\sin(\arctan x)=\sqrt{\frac{\tan^2(\arctan x)}{\tan^2(\arctan x)+1}}
$$
the answer is $\frac{x}{\sqrt{x^2+1}}+C$.
My Question: Is there another way to find this integral without using trigonometry?

Let $u=x^2+1$.
$$I=\frac{1}{2}\int{u^{-\frac{3}{2}}(u-1)^{-\frac{1}{2}}}du=\frac{1}{2}\int{u^{-2}\left(1-\frac{1}{u}\right)^{-\frac{1}{2}}}du$$
Using the general formula $\int{(f(x))^nf'(x)}dx=\dfrac{(f(x))^{n+1}}{n+1}+c$ (which can be verified by differentiating the right side),
$$I=\left(1-\frac{1}{u}\right)^{\frac{1}{2}}+c=\frac{x}{\sqrt{x^2+1}}+c$$
