Monday, January 3, 2022

12 Days of Christmas Integrals: ∫ x^2 ∙ exp(x)^2 dx

 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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