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?
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 $$\int \frac{u \,du}{(1 + u^2)^{3 / 2}} .$$
$$\int\frac{1}{(x^2+1)\sqrt{x^2+1}}dx\overset{x=\frac1y}= -\int \frac y{(1+y^2)^{3/2}}dy=\frac1{\sqrt{1+y^2}}+C $$
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 $\frac{x}{\sqrt{1+x^2}}$.
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),
$$I=\left(1-\frac{1}{u}\right)^{\frac{1}{2}}+c=\frac{x}{\sqrt{x^2+1}}+c$$
You can use hyperbolic trigonometry. Set $x=\sinh t$, $x^2+1=\cosh^2 t$, $dx=\cosh t dt$, and you get:
$$\begin{array}{rcl}\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}&=&\int\frac{\cosh t\,dt}{\cosh^3 t}\\&=&\int\frac{dt}{\cosh^2 t}\\&=&\tanh t+C\\&=&\frac{\sinh t}{\cosh t}+C\\&=&\frac{x}{\sqrt{x^2+1}}+C\end{array}$$
Substitute $x = \sqrt{\dfrac{1-y}{1+y}}$ :
$$\begin{align*} & \int \frac{dx}{\left(x^2+1\right) \sqrt{x^2+1}} \\ &= - \int \frac{\sqrt{\frac{1+y}{1-y}} \, \frac{dy}{(1+y)^2}}{\left(\frac{1-y}{1+y} + 1\right) \sqrt{\frac{1-y}{1+y} + 1}} \\ &= - \frac1{2\sqrt2} \int \frac{dy}{\sqrt{1-y}} \\ &= \sqrt{\frac{1-y}2} + C \\ &= \frac x{\sqrt{x^2+1}} + C \end{align*}$$
