Question Number 150007 by mathdanisur last updated on 08/Aug/21

∫ ((x^2 + x)/(x^6 + 1)) dx = ?

Answered by Ar Brandon last updated on 08/Aug/21

I=∫((x^2 +x)/(x^6 +1))dx=∫(x^2 /(x^6 +1))dx+∫(x/(x^6 +1))dx =(1/3)∫((d(x^3 ))/((x^3 )^2 +1))+(1/2)∫((d(x^2 ))/((x^2 )^3 +1)) =((tan^(−1) (x^3 ))/3)+(1/2)∫((d(x^2 ))/((x^2 +1)(x^4 −x^2 +1))) =((tan^(−1) (x^3 ))/3)+(1/2)∫((1/3)∙(1/(x^2 +1))−(1/3)∙((x^2 −2)/(x^4 −x^2 +1)))dx^2 =((tan^(−1) (x^3 ))/3)+(1/6)ln(x^2 +1)−(1/(12))ln(x^4 −x^2 +1)+(1/(2(√3)))tan^(−1) (((2x^2 −1)/( (√3))))+C =((tan^(−1) (x^3 ))/3)+(1/(12))ln∣((x^4 +2x^2 +1)/(x^4 −x^2 +1))∣+ ((√3)/6)tan^(−1) (((2x^2 −1)/( (√3))))+C

Commented bymathdanisur last updated on 08/Aug/21

Thank you Ser