How do you integrate $int x^c e^(x^d)dx$ , where $c>d$ , using integration by parts?

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Answer 1 · 2016-10-04T08:27:17

Use the recurrence formula

$I_(d+k)+(k+1)/dI_k = 1/dx^(k+1)e^(x^d)$

with initial $k = c-d$ and

$I_(d-1)= 1/de^(x^d)$

We use $n, m$ instead of $c,d$ ,
.

$d/(dx)(x^n e^(x^m)) = nx^(n-1)e^m+ m x^(n+m-1) e^(x^m)$

Calling now

$I_n = int x^n e^(x^m)dx$ we have the recurrence relationship

$mI_(n+m-1)+nI_(n-1)=x^n e^(x^m)$

with

$I_(m-1)=int x^(m-1)e^(x^m)dx = 1/me^(x^m)$ or calling $k = n-1$

$mI_(m+k)+(k+1)I_k = x^(k+1)e^(x^m)$

Finally

$I_(m+k)+(k+1)/mI_k = 1/mx^(k+1)e^(x^m)$

Example. $n=5, m=3$ so $k=2$ and we want

$I_(m+2)= I_5 = int x^5 e^(x^3)dx$

we know that $I_(m-1) = I_2$ and

$I_2=1/3e^(x^3)$ so

$I_(m+2)+(2+1)/mI_2=1/mx^3e^(x^3)$ so

$I_5 +3/3I_2=1/3x^3e^(x^3)$ or

$I_5 = 1/3x^3 e^(x^3)-1/3e^(x^3)=1/3(x^3-1)e^(x^3)$

How do you integrate int x^c e^(x^d)dxint x^c e^(x^d)dx, where c>dc>d, using integration by parts?