Question

How do you find the integral `ln((x^2)-1)`?

Answer 1

Refer to explanation

Explanation:

Using integration by parts we have that

`int ln(x^2-1)*x'dx=x*ln(x^2-1)-intx*(2x)/(x^2-1)dx= x*ln(x^2-1)-2intx^2/(x^2-1)dx`

Hence we have the integral `intx^2/(x^2-1)dx` which can be written

as follows

`intx^2/(x^2-1)dx=int((x^2-1)+1)/(x^2-1)dx=int(1+1/(x^2-1))dx=x+int1/(x^2-1)dx=x+1/2*int(1/(x-1)-1/(x+1))=x+1/2(ln(x-1)-ln(x+1))`

Finally we get

`int ln(x^2-1)*dx=x(ln(x^2-1)-2)-ln(1-x)+ln(1+x)`