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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