Sunday, December 26, 2021

12 Days of Christmas Integrals: ∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx

 12 Days of Christmas Integrals:  ∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx


On the Second day of Christmas Integrals, the integral featured today is...


∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx


This integral can be approached by using the substitution method:  


Let u = arctan(x^2).  Then:


du = 1 ÷ (1 + x^4) ∙  d/dx(x^2)

du = 1 ÷ (1 + x^4) ∙  (2  ∙ x) dx

1/2 du = x ÷ (1 + x^4) dx


Then:


∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx


= ∫ u ∙ 1/2 du


= 1/2 ∙ ∫ u du


 = u^4/4 + C


Substitute back:


= (arctan(x^2))^2 / 4 + C



Eddie 


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