Integrate $\int \frac{ e^{\tan^{-1}x}}{(1+x^2)^2} dx$

Question

How do I carry out the integration $$\int \frac{ e^{\tan^{-1}x}}{(1+x^2)^2} dx$$

The answer, from Wolfram Alpha is $$e^{\tan^{-1}x}\left ( \frac 2 5 + \frac {2x+1} {5(x^2+1)} \right) $$

Answer 1

Let $\displaystyle I=\int\frac{e^{\tan^{-1}x}}{(1+x^2)^2}dx$, and integrate by parts with $\displaystyle u=\frac{1}{x^2+1}$ and $\displaystyle dv=\frac{e^{\tan^{-1}x}}{1+x^2}dx$ to get

$\hspace{.8 in}\displaystyle I=\frac{e^{\tan^{-1}x}}{x^2+1}+2\int\frac{x e^{\tan^{-1}x}}{(1+x^2)^2}dx$. $\;\;\;(1)$

Now integrate by parts with $\displaystyle u=\frac{x}{x^2+1}$ and $\displaystyle dv=\frac{e^{\tan^{-1}x}}{1+x^2}dx$ to get

$\;\;\;\displaystyle \int\frac{x e^{\tan^{-1}x}}{(1+x^2)^2}dx=\frac{x e^{\tan^{-1}x}}{x^2+1}-\int\frac{e^{\tan^{-1}x}}{(1+x^2)^2}dx+\int\frac{x^2 e^{\tan^{-1}x}}{(1+x^2)^2}dx $

$\displaystyle\hspace{1.3 in}=\frac{x e^{\tan^{-1}x}}{x^2+1}-\int\frac{e^{\tan^{-1}x}}{(1+x^2)^2}dx+\int\frac{e^{\tan^{-1}x}}{1+x^2}dx-\int\frac{e^{\tan^{-1}x}}{(1+x^2)^2}dx$

$\displaystyle\hspace{1.3 in}=\frac{x e^{\tan^{-1}x}}{x^2+1}-2I+e^{\tan^{-1}x}$

Substituting in (1) gives

$\displaystyle\hspace{1.3 in}I=\frac{e^{\tan^{-1}x}}{x^2+1}+\frac{2x e^{\tan^{-1}x}}{x^2+1}-4I+2e^{\tan^{-1}x}$,

so $\displaystyle\hspace{1.1 in}I=\frac{1}{5}\left[\frac{e^{\tan^{-1}x}}{x^2+1}+\frac{2x e^{\tan^{-1}x}}{x^2+1}+2e^{\tan^{-1}x}\right]+C$

$\displaystyle\hspace{1.4 in}=\frac{1}{5}e^{\tan^{-1}x}\left[\frac{2x+1}{x^2+1}+2\right]+C$.

Answer 2

Notice that $\tan^{-1}(x)$ and $1/(1+x^2)$ appear in your integrand. This means a $u$-substitution with $\tan^{-1}(x)$ could be useful.

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

Let $u=\tan^{-1}(x)$ then $du = \frac{1}{1+x^2}dx$ and $x = \tan(u)$. So the integral becomes:

$$\int \frac{e^{\tan^{-1}(x)}}{(1+x^2)^2}dx=\int \frac{e^{u}}{1+\tan^2(u)} du$$

Recall that $1+\tan^2(x) = \sec^2(x) = 1/\cos^2(x)$ so

$$\int \frac{e^{u}}{1+\tan^2(u)} du=\int e^{u} \cos^2(u) du$$

Finally $\cos^2(u) = \frac12 (1+\cos(2u))$ and

$$\int e^{u} \cos^2(u) du=\frac12 \int (1+\cos(2u)) e^u du$$

Answer 3

$$\int\frac{e^{\tan^-1(x)}}{(1+x^2)dx}=\int{e^{\arctan(x)}}\cdot\frac{dx}{1+x^2}\cdot\frac{1}{1+x^2}$$ $$=\left|\arctan x=t\Rightarrow \frac{dx}{1+x^2}=dt; x=\tan t\Rightarrow 1+x^2=1+\tan^2 t=\frac{1}{\cos^2 t}\right|$$ $$=\int{e^t}\cdot dt\cdot\frac{1}{\frac{1}{\cos^2 t}}=\int e^t\cos^2 t dt=\int e^t\left(\frac{1+\cos 2t}{2}\right)dx=$$ $$=\frac{1}{2}\int e^t(1+\cos 2t)dt=\frac{1}{2}\int e^t dt+\frac{1}{2}\int e^t\cos 2tdt=\frac{e^t}{2}+\frac{1}{2}I$$ Now we have a solution of the integral: $$I=\int e^t\cos 2tdt=\left|u=e^t\Rightarrow du=e^tdt; dv=\cos 2t\Rightarrow v=\frac{\sin 2t}{2}\right|=$$ $$=\frac{e^t\sin 2t}{2}-\frac{1}{2}\int e^t\sin 2t dt=\frac{e^t\sin 2t}{2}-\frac{1}{2}I_1$$ Now solution of the integral $I_1$ $$I_1=\int e^t\sin 2tdt=\left|u=e^t\Rightarrow du=e^tdt; dv=\sin 2t\Rightarrow v=-\frac{\cos 2t}{2}\right|=$$ $$=-\frac{e^t\cos 2t}{2}+\frac{1}{2}\int e^t\cos 2t dt$$ From here we have: $$\int e^t\cos 2t dt=\frac{e^t \sin 2t}{2}-\frac{1}{2}\left(-\frac{e^t \cos 2t}{2}+\frac{1}{2}\int e^t\cos 2t dt\right)$$ $$\frac{5}{4}\int e^t \cos2t dt=\frac{e^t \sin 2t}{2}+\frac{e^t \cos 2t}{4}$$ $$I=\int e^t \cos2t dt=\frac{2e^t \sin 2t}{5}+\frac{e^t \cos 2t}{5}$$ Finally we have: $$\int\frac{e^{\arctan x dx}}{(1+x^2)2}dx=\frac{e^t}{2}+\frac{e^t\sin 2t}{5}+\frac{e^t\cos 2t}{10}$$ $$=e^t\left(\frac{1}{2}+\frac{\sin 2t}{5}+\frac{\cos 2t}{10}\right)=e^t\cdot\frac{5+2\sin 2t+\cos 2t}{10}$$ $$=e^t\cdot\frac{5+2\frac{2\tan t}{1+\tan^2 t}+\frac{1-\tan^2 t}{1+\tan^2 t}}{10}$$ $$=e^{\arctan x}\cdot\frac{5+5\tan^2 t+4\tan t+1-\tan^2 t}{10(1+\tan^2 t)}$$ $$=e^{\arctan x}\cdot\frac{6+4\tan t+4\tan^2 t}{10(1+\tan^2 t)}=e^{\arctan x}\cdot\frac{3+2\tan t+2\tan^2 t}{5(1+\tan^2 t)}$$ $$=e^{\arctan x}\cdot\frac{3+2x+2x^2}{5(1+x^2)}+C$$ I hope you help.