Q.

The integral (x8-x2)dx(x12+3x6+1)tan-1(x3+1x3) is equal to:         [2024]

1 loge(|tan-1(x3+1x3)|)+C  
2 loge(|tan-1(x3+1x3)|)3+C  
3 loge(|tan-1(x3+1x3)|)13+C  
4 loge(|tan-1(x3+1x3)|)12+C  

Ans.

(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=13duu=ln|u|3=ln(|tan-1(x3+1x3)|)13+C