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?

$$\int\frac{1}{(x^2+1)\sqrt{x^2+1}}dx\overset{x=\frac1y}=
-\int \frac y{(1+y^2)^{3/2}}dy=\frac1{\sqrt{1+y^2}}+C
$$
