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
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