Question

`int1/((x)(6(logx)^2 +7logx + 2)) `dx

Answer 1

`I = int 1/(x(6logx)^2+7logx+2)dx`
Let `logx = t,` then, `dx/x = dt`
`:. I = int 1/(6t^2+7t+2) dt`
Now, `6t^2+7t+2 = 6t^2+4t+3t+2 = (3t+2)(2t+1)`
`:. I = int 1/((3t+2)(2t+1))dt`
Now, `1/((3t+2)(2t+1)) = A/(2t+1)+B/(3t+2)`
If we put, `t = -1/2, A = 1/(3(-1/2)+2)=> A = 2`
If we put, `t = -2/3, B = 1/(2(-2/3)+1) =>B = -3`
`:. I = int 2/(2t+1)dt +int (-3)/(3t+2)dt`
`=>I = 2(log(2t+1)/2) - 3(log(3t+2)/3)+c`
`=>I = log(2t+1)-log(3t+2)+c`
`=>I = log((2t+1)/(3t+2))+c`
`=>I = log((2logx+1)/(3logx+2))+c`