I'm stuck on a question Calculus:Integration

Question

$$\int \frac{e^x (x^4 + 4)}{(x^2+1)^{5/2}} \,\mathrm{d}x$$

I have tried a lot of methods but none is able to simplify the expression, it keep s getting more complex and harder to solve, I would be glad if someone would be willing to give me a kickstart, Thanks

I tried to simplify the expression into elementary form by opening the brackets and multiplying but nothing simplified, can a substitution be done to simplify?

P.S. This is not for homework I am a 12th-grade kid and I am learning calculus on my own, I was stuck on this question I found online.

Answer 1

I may contribute here.. This may be more simpler ..

We can substitute $x = \tan(t), \implies dx = \sec^{2}(t) dt $ and get $$ \int \frac{e^{\tan(t)}(\tan^{4}(t) + 4)}{(\tan^{2}(t) + 1)^{5/2}}dx = \int \frac{e^{\tan(t)}(\tan^{4}(t) + 4)}{(\sec^{2}(t))^{5/2}} \sec^{2}(t) dt $$ $$ = \int \frac{e^{\tan(t)}(\tan^{4}(t) + 4)}{\sec^{3}(t)} dt $$

Now remember that

$$ e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} $$

so we have

$$ e^{\tan(t)} = \sum_{n=0}^{\infty} \frac{\tan^{n}(t)}{n!} $$

Using this for the integral, we calculate instead :

$$ \int \sum_{n=0}^{\infty} \frac{\tan^{n}(t)}{n!} \frac{(\tan^{4}(t) + 4)}{\sec^{3}(t)} dt $$ $$ = \int \cos^{3}(t) \sum_{n=0}^{\infty} \frac{\tan^{n+4}(t) + 4 \tan^{n}(t)}{n!} dt $$ $$ = \sum_{n=0}^{\infty} \frac{1}{n!}\left[ \int \cos^{3}(t) \left( \tan^{n+4}(t) + 4 \tan^{n}(t) \right) dt \right] $$

I think this may be simpler, but I have not found full solution. Maybe it can be solved directly, or maybe using induction. Hope this helps.