Question

How do you evaluate the integral of `intln(x^2 - 1)dx`?

Answer 1

Use integration by parts and integration by partial fractions to find that the integral evaluates to

`xln(x^2-1)-2x-ln|x-1|+ln|x+1|+C`

Explanation:

We will proceed using integration by parts and partial fraction decomposition .

We start by using integration by parts.

Let `u = ln(x^2-1)` and `dv = dx`

Then `du = (2x)/(x^2-1)dx` and `v = x`

By the integration by parts formula `intudv = uv - intvdu`

`intln(x^2-1)dx = xln(x^2-1) - int(2x^2)/(x^2-1)dx`

`=xln(x^2-1)-2int(1+1/(x^2-1))dx`

`=xln(x^2-1)-2int1dx -2int1/(x^2-1)dx`

`=xln(x^2-1)-2x-2int1/((x+1)(x-1))dx`

To evaluate the remaining integral, we can use partial fractions.

`1/((x+1)(x-1))=A/(x+1)+B/(x-1)`

`=> 1 = A(x-1) + B(x+1)`

`=(A+B)x + (-A+B)`

`=>{(A+B=0), (-A+B=1):}`

`=>{(A = -1/2),(B=1/2):}`

`=>int(1/((x+1)(x-1))dx = int((1/2)/(x-1)-(1/2)/(x+1))dx`

`=1/2(int1/(x-1)dx-int1/(x+1)dx)`

`=1/2(ln|x-1|-ln|x+1|)+C`

Substituting this back into the prior result, we have

`intln(x^2-1)dx =`
`= xln(x^2-1)-2x-ln|x-1|+ln|x+1|+C`