General method to find $f(x)$ in $\int e^x(f(x)+f'(x))$ It is known that
$$\int e^x\left(f(x) + f'(x)\right) \,dx=  e^xf(x)$$
which can be verified using Int by parts.
I wanted to know whether there was a general procedure to find the $f(x)$ here, or we are left up to our intuition?
Edit:
Consider the following:
$$\int e^x\left(\dfrac{1+nx^{n-1}-x^{2n}}{(1-x^n)\sqrt{1-x^{2n}}}\right)\,dx$$
Now all we have to do is determine the pair $f+f'$. How do we do it? Is there a general method to approach these type of questions?
 A: This is pretty straightforward with your observation above. Multiply both sides by $e^x$ to get that
$$e^xg(x) = e^x(f(x) +f'(x)) = (e^xf(x)) '. $$
Integrate both sides and we have 
$$\int e^xg(x) \, dx = e^x f(x) $$
and so 
$$f(x) = e^{-x} \int e^x g(x). $$
A: Suppose there was a general technique which, given some function $g$, produced a function $f$ such that $f+f'=g$.
Then, given any arbitrary function $h$, we could apply this technique to the function $e^{-x}h(x)$, giving a function $k$ such that
$$
k(x)+k'(x)=e^{-x}h(x)
$$
But then we would have
$$
e^{x}k(x)+e^xk'(x)=h(x)\\
(e^xk(x))'=h(x)
$$
That is, given any function $h$, we could produce a function whose derivative was $h$. So any such technique could be used to integrate arbitrary functions. Since symbolic integration of arbitrary functions is impossible, no such technique can exist. The best you can hope for is that you get lucky in some specific cases.
A: I think there is no general method ,we have to do some observation and manipulation
$$\int e^x\left(\dfrac{(1-x^n)(1+x^n)+nx^{n-1}}{(1-x^n)\sqrt{1-x^{2n}}}\right)\,dx$$
$$\int e^x\left(\dfrac{(1+x^n)+nx^{n-1}}{\sqrt{1-x^{2n}}}\right)dx$$
$$\int e^x\left(\frac{(1+x^n)}{\sqrt{1-x^{2n}}}+\frac{nx^{n-1}}{{(1-x^n)\sqrt{1-x^{2n}}}}\right)dx$$
$$\int e^x\left(f(x) +f'(x)\right)dx$$
