How to integrate $\int\frac{xe^{tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx$?

Question

Original question: (updated question in section EDIT)

How to evaluate the following integral?

$$\int\frac{1}{(1+x^2)^\frac{3}{2}}\,dx$$

I tried substitution:
$$ t = 1+x^2 = \varphi \\ \frac{\,dt}{\,dx} = \varphi' = 2x \implies \,dt = 2x\,dx$$

... this doesn't help me, then I thought, maybe this way:
$$ t = 1+x^2 \implies x = \sqrt{t - 1} = \varphi \\ \frac{\,dx}{\,dt} = \varphi' = \frac{1}{2\sqrt{t-1}} \implies \,dx = \frac{1}{2\sqrt{t-1}}\,dt$$

so I have:

$$\int\frac{1}{(1+x^2)^\frac{3}{2}} \,dx= \frac{1}{2}\int\frac{\,dt}{t^{\frac{3}{2}}\sqrt{t-1}} = \frac{1}{2}\int\frac{\,dt}{\sqrt{t^4-t^3}}$$

which doesn't look very promising.

Then I thought: $$ t = x^2 \implies x = \sqrt{t} = \varphi \\ \frac{\,dx}{\,dt} = \varphi' = \frac{1}{2\sqrt{t}} \implies \,dx = \frac{\,dt}{2\sqrt{t}}$$

which gives me:

$$\int\frac{1}{(1+x^2)^\frac{3}{2}} \,dx= \frac{1}{2}\int\frac{\,dt}{(1+t)^{\frac{3}{2}}\sqrt{t}}$$

The answer looks like someone could just "guess it": http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ea57hoiergo but I don't know how to "get there" by calculation.

EDIT:

The actual problem was the following integral:

$$\int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx$$

and my idea was to apply partial integration i.e. $\int u\,dv = uv - \int v\,du$ where $$ u = e^{\tan^{-1}x} \implies \,du = e^{\tan^{-1}x} \cdot \frac{1}{1 + x^2} \,dx = \frac{e^{\tan^{-1}x}}{1+x^2} \,dx \\ \,dv = \frac{x}{(1+x^2)^\frac{3}{2}} \implies v = -\frac{1}{\sqrt{1+x^2}}$$

because $$t = 1 + x^2 = \varphi \implies \varphi' = 0 + 2x = 2x \\ \frac{\,dt}{\,dx} = \varphi' = 2x \implies \,dt = 2x\,dx \\ \int \frac{\frac{1}{2}\,dt}{t^\frac{3}{2}} = \frac{1}{2} \int\frac{\,dt}{t^\frac{3}{2}} = \frac{1}{2} \cdot \frac{t^{-\frac{1}{2}}}{-\frac{1}{2}} + C = -t^{-\frac{1}{2}} + C = -\frac{1}{\sqrt{1+x^2}} + C$$

and now we have:

$$\int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} - \int \bigg( -\frac{1}{\sqrt{1+x^2}} \bigg) \cdot \frac{e^{\tan^{-1}x}}{1+x^2}\,dx \\ = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} + \int \frac{e^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx$$

now I want to solve $\int \frac{e^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx$ and my approach was the following: $$ u = e^{\tan^{-1}x} \implies \,du = \frac{e^{\tan^{-1}x}}{1+x^2} \,dx \\ \,dv = \frac{1}{(1+x^2)^\frac{3}{2}}\,dx \implies v = ?$$

and this is the part I couldn't figure out i.e. $\int\frac{1}{(1+x^2)^\frac{3}{2}}\,dx$ but as you all pointed out, we can solve this by using trig (as shown here: Integrate Form $du / (a^2 + u^2)^{3/2}$) or hyperbolic (as shown in the answer provided by georg) substitutions which gives us:

$$\int\frac{1}{(1+x^2)^\frac{3}{2}}\,dx = \cdots = \frac{x}{\sqrt{1+x^2}} + C \implies v = \frac{x}{\sqrt{1+x^2}} $$

so we finally get:

$$\int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} + \int \frac{e^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx \\ = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} + \frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}} - \int \frac{x}{\sqrt{1+x^2}}\cdot \frac{e^{\tan^{-1}x}}{1+x^2}\,dx \\ = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} + \frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}} - \int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx \\ \implies \\ 2 \int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} + \frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}} \\ \int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx = \frac{(x-1)e^{\tan^{-1}x}}{2\sqrt{1+x^2}} + C$$

which is the answer to the problem.

I also noticed a different approach:

$$\int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx = \int\frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}(1+x^2)}\,dx \\ t = \tan^{-1}x = \varphi \implies \varphi' = \frac{1}{1+x^2} \\ \frac{dt}{dx} = \varphi' \implies \,dt = \frac{\,dx}{1+x^2} \\ \int\frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}(1+x^2)}\,dx = \int\frac{\tan(t) e^t \,dt}{\sqrt{1 + \tan^2(t)}} = \int\frac{\tan(t) e^t \,dt}{\sqrt{\frac{1}{\cos^2(t)}}} = \int\sin(t) e^t \,dt \\ u = \sin t \implies \,du = \cos t\,dt \\ \,dv = e^t \,dt \implies v = e^t \\ \int\sin(t) e^t \,dt = \sin (t)e^t - \int e^t\cos(t)\,dt$$

now we solve $\int e^t\cos(t)\,dt$ : $$ u = \cos t \implies \,du = -\sin t \,dt \\ \,dv = e^t\,dt \implies v = e^t \\ \int e^t\cos(t)\,dt = \cos(t)e^t - \int(-\sin(t))e^t\,dt$$

so we finally have:

$$ \int\sin(t) e^t \,dt = \sin(t)e^t - \bigg(\cos(t)e^t - \int(-\sin(t))e^t\,dt \bigg) \implies \\ 2\int\sin(t) e^t \,dt = \sin(t)e^t - \cos(t)e^t \implies \int\sin(t) e^t \,dt = \frac{(\sin(t) - \cos(t))e^t}{2} \\ \int\sin(\tan^{-1}x) e^{\tan^{-1}x} \,dx = \frac{(\sin(\tan^{-1}x) - \cos(\tan^{-1}x))e^{\tan^{-1}x}}{2} + C $$

and since: $$\sin(\tan^{-1}x) = \frac{x}{\sqrt{1+x^2}} ,\, \cos(\tan^{-1}x) = \frac{1}{\sqrt{1+x^2}}$$

we have: $$ \int\sin(\tan^{-1}x) e^{\tan^{-1}x} \,dx = \frac{(x-1)e^{\tan^{-1}x}}{2\sqrt{1+x^2}} + C$$

and that's the answer.

Answer 1

Substitution: $x = \sinh u$, $dx = \cosh u\,du$

\begin{align*} \int\frac{1}{(1+x^2)^\frac{3}{2}}\,dx & =\int\frac{\cosh u}{(1+\sinh^2 u)^\frac{3}{2}}\,du\\ &= \int\frac{\cosh u}{(\cosh^2 u)^\frac{3}{2}}\,du\\ &=\int\frac{1}{\cosh^2 u}\,du\\ &=\tanh u + C\\ &=\frac{x}{\sqrt{1+x^2}} + C \end{align*}