12 Days of Christmas Integrals: ∫ x^2 ∙ exp(x)^2 dx
On the Tenth day of Christmas Integrals, the integral featured today is...
∫ x^2 ∙ exp(x)^2 dx
exp(x) = e^x
Start with the substitution:
u = exp(x)
du = exp(x) dx
1 ÷ exp(x) du = dx
1 ÷ u du = dx
u^2 = exp(x)^2
ln(u) = x
ln(u)^2 = x^2
Then:
∫ x^2 ∙ exp(x)^2 dx
= ∫ ln(u)^2 ∙ u^2 ∙ 1/u du
= ∫ u ∙ ln(u)^2 du
Note: We did this integral on the Seventh Day of 12 Days of Christmas integrals (December 31, 2021). Check out that post for details!
= u^2/2 ∙ ln(u)^2 - u^2/2 ∙ ln(u) + u^4/4 + C
Recall u = exp(x):
= 1/2 ∙ (exp(x))^2 ∙ x^2 - 1/2 ∙ (exp(x))^2 ∙ x + 1/4 ∙ (exp(x))^2 + C
Eddie
All original content copyright, © 2011-2022. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
