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?
Best Answer
Hint One method is to apply the Euler substitution, $$\sqrt{1 + x^2} = t + x , \qquad dx = - \frac{1 + t^2}{2 t^2}\,dt ,$$ which transforms the integral to $$-4 \int \frac{t\,dt}{(1 + t^2)^2}.$$
Another is to apply the substitution $$x = \frac{1}{u}, \qquad dx = -\frac{du}{u^2},$$ which transforms the integral to $$\int \frac{u \,du}{(1 + u^2)^{3 / 2}} .$$