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?

Another way is to use power series:
$$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n} x^n \tag{1} $$
$$\text{by differentiation and reindexing}\quad \frac{1}{(1-x)\sqrt{1-x}} = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}(2n+1) x^n \tag{2}$$
$$\text{by }x\mapsto -x^2\quad \frac{1}{(1+x^2)\sqrt{1+x^2}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}(2n+1)(-1)^{n} x^{2n} \tag{3}$$
$$\text{by termwise integration}\quad \int_{0}^{x}\frac{dt}{(1+t^2)\sqrt{1+t^2}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}(-1)^n x^{2n+1}\tag{4} $$
and by comparing $(4)$ and $(1)$ we have that the RHS of $(4)$ is $\frac{x}{\sqrt{1+x^2}}$.
