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?

You can use hyperbolic trigonometry. Set $x=\sinh t$, $x^2+1=\cosh^2 t$, $dx=\cosh t dt$, and you get:
$$\begin{array}{rcl}\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}&=&\int\frac{\cosh t\,dt}{\cosh^3 t}\\&=&\int\frac{dt}{\cosh^2 t}\\&=&\tanh t+C\\&=&\frac{\sinh t}{\cosh t}+C\\&=&\frac{x}{\sqrt{x^2+1}}+C\end{array}$$
