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

Question

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

Charles Arocha · Accepted Answer

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)# Making a call right 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)#

Emily Ammerman · Answer

undefined To integrate $ \int x^c e^{x^d} \, dx $ where $ c > d $ using integration by parts, we let:

$ u = x^c $ and $ dv = e^{x^d} \, dx $.

Then, we find:

$ du = cx^{c-1} \, dx $ and $ v = \int e^{x^d} \, dx $.

Since $ v $ does not have a simple antiderivative, we cannot directly integrate it. Therefore, integration by parts would not be applicable in this case.

How do you integrate #int x^c e^(x^d)dx#, where #c&gt;d#, using integration by parts?

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