Integration of $\int{\frac{e^{\arctan{x}}}{\sqrt{1 + x^2}}dx}$

Question

I am not familiar with integral a lot but I know most of important rule.
$$\int{\frac{e^{\arctan{x}}}{\sqrt{1 + x^2}}dx}$$

My method :

I assume $u = \arctan{x}$ so $\displaystyle du = \frac{1}{1 + x^2}dx$ and I get this integral:
$$\int{\sec{u}e^{u}du}$$

But I dont have any more idea!

Can you please help me? Thanks.

score 0 · Answer 1 · answered Nov 06 '17 at 07:57

0

Choose $x=\tan u$, so $dx=\frac{\mathrm{d}u}{\cos^2(u)}$ and your integral becomes

$$\int \frac{e^u}{\cos^2(u)\sqrt{1+\tan^2(u)}}\mathrm{d}u $$

then recall that $1+\tan^2(u)=\frac{1}{\cos^2(u)}$, so you have

$$\int\frac{e^u}{\cos(u)}\mathrm{d}u $$

Wolfram Alpha gives a closed form of this in term of the Hypergeometric function, I doubt a prettier closed form can be found

answered Nov 06 '17 at 07:57

user438666

Maybe. Because I cant solve it too. – Amin Nov 06 '17 at 07:59
Why can't we do integration by parts on the product terms? – learning Nov 06 '17 at 08:00
How would you do that? Differentiating $\frac{1}{\cos(x)}$ isn't going to make the integral easier, and differentiating $e^u$ and using the primitive of $\frac{1}{\cos(x)}$ neither. – user438666 Nov 06 '17 at 08:03
Wouldn't differentiating $e^u$ give back $\int e^u sec(u) du$? – learning Nov 06 '17 at 08:05
@Amin, sorry, I somehow misread and didn't realize the form I arrived at is exactly the one you arrived at, not that useful I guess – user438666 Nov 06 '17 at 08:05
@learning you would have $\int e^u\sec(u)\mathrm{d}u=e^u \sec(u) -\int e^u\frac{1+\sin(u)}{\cos(u)}\mathrm{d}u$ which isn't any easier that what we started from – user438666 Nov 06 '17 at 08:07
Whoops, never mind. Thanks! – learning Nov 06 '17 at 08:08

1 Answers1