`int(ln((x-1)/(x+1)))/(x^2-1)dxi se q u a lto` `1/2(ln((x-1)/(x+1)))^2+C` `1/2(ln((x+1)/(x-1)))^2+C` `1/4(ln((x-1)/(x+1)))^2+C` `1/4(ln((x+1)/(x-1)))^2+C`

Options

A. `(1)/(2)("In"((x-1)/(x+1)))^(2)+C`
B. `(1)/(2)("In"((x+1)/(x-1)))^(2)+C`
C. `(1)/(4)("In"((x-1)/(x+1)))^(2)+C`
D. `(1)/(4)("In"((x+1)/(x-1)))`

Answer

C

Explanation

`I=int("In"((x-1)/(x+1)))/(x^(2)-1)dx`
Let `t="In"((x-1)/(x+1))`
or `(dt)/(dx)=(x+1)/(x-1){(x+1-(x-1))/((x+1)^(2))}=(2)/((x^(2)-1))`
or `(dx)/(x^(2)-1)=(dt)/(2)`
` :. I=(1)/(2)int t dt=(1)/(4)t^(2)+C=(1)/(4)("In"((x-1)/(x+1)))^(2)+C`