Generalized Infinite Integration by Parts

Question

During my studies in calc 2, I became fascinated by the integral $\int e^{-x^2}dx$ after hearing from the professor that it has no elementary function as its integral. I came up with an interesting technique to try the integral by using Integration by Parts like so: $$\int e^{-x^2}dx=xe^{-x^2}+2\int x^2e^{-x^2}dx $$ $$\int e^{-x^2}dx=xe^{-x^2}+\frac {2} {3}x^3e^{-x^2}+\int x^4e^{-x^2}dx$$ and continuing this until the pattern became obvious and I came up with the following equation: $$\int e^{-x^2}dx=e^{-x^2}\sum_{i=0}^{\infty} \frac {2^i} {(2i+1)!!}x^{2i+1}+C$$ where $(2i+1)!!$ denotes the double factorial $(2n+1)(2n-1)(2n-3)...(2 \text{ or } 1)$. From here, I thought about a generalized case for any infinitely differentiable function $f(x)$ and for an argument raised to any power $n$. $$\int f(x^n)dx $$ $$u=f(x^n), du=nx^{n-1}f'(x^n), dv=dx, v=x$$ $$\int f(x^n)dx=xf(x^n)-n\int x^nf'(x^n)dx$$ and eventually: $$=\sum_{i=0}^{\infty}\frac {(-n)^{i}x^{ni+1}} {(ni+1)(n(i-1)+1)...(n+1)(1)}f^{(i)}(x^n)+C$$ My question is twofold. First, did I make any glaring mistakes, and second, is this particular formula useful or novel?

Answer 1

Please check your $du$ again in $$ u=f(x^n), du=nx^{n-1}f(x^n), dv=dx, v=x$$

Also check this formula $$ \int f(x^n)dx=xf(x^n)-n\int x^nf(x^n)dx$$

What is the advantage of your method in comparison with integrating the Taylor series of a function?