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

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

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## marked as duplicate by Davide Giraudo, Alexander Gruber♦May 10 at 17:04

Have you tried trig substitutions –  mathguy May 10 at 16:25
Hint: Look up some well known hyperbolic function identities. –  Raj May 10 at 16:26
If you know how to solve: $$f(x,y)=\int{\frac {1}{\sqrt {{y}^{2}+{x}^{2}}}}\,dx$$ then you can use differentiation to get to your case because: $$\left(\frac{\partial}{\partial y} f(x,y)\right)_{y=-1}=\int{\frac{1}{ \left( {x}^{2}+1 \right) ^{3/2}}}\,dx$$ The point being that you can use differentiation acting on some parameter you arbitrarily introduce, to increase the order of the exponent in problems from one you know how to solve to one you don't. In this case from $1/2$ to $3/2$. This assumes you can tackle the reduced exponent case by some other means. –  Graham Hesketh May 10 at 17:07

$\int\frac{1}{(1+x^2)^\frac{3}{2}}dx=\int\frac{coshu}{(1+sinh^2u)^\frac{3}{2}}du= \int\frac{coshu}{(cosh^2u)^\frac{3}{2}}du= \int\frac{coshu}{(cosh^2u)^\frac{3}{2}}du=$
$=\int\frac{1}{cosh^2u}du=tanhu=\frac{x}{\sqrt{1+x^2}} + C$