The integral ∫(x8-x2)dx(x12+3x6+1)tan-1(x3+1x3) is equal to: [2024]
(3)
Let I=∫(x8-x2) dx(x12+3x6+1)tan-1(x3+1x3)
Put tan-1(x3+1x3)=u
⇒11+(x3+1x3)2·3(x2-1x4)dx=du
⇒(x6x12+3x6+1×x6-1x4)dx=du3
⇒I=13∫duu=ln|u|3=ln(|tan-1(x3+1x3)|)13+C