Question

How do you evaluate the definite integral `int x/(x^2-1)` from `[2,3]`?

Answer 1

1/2ln(8/3)

Explanation:

`intxdx/(x^2-1)=intxdx/((x+1)(x-1))`
`x/((x+1)(x-1))=A/(x+1)+B/(x-1)`
`x=A(x-1)+B(x+1)`
`x=1 =>1=0+2B ; B=1/2`
`x=-1 =>-1=0-2A ; A=1/2`
`I=intxdx/(x^2-1)=1/2(int1dx/(x+1)+int1dx/(x-1))=1/2(ln(x+1)+ln(x-1))=1/2(ln(x+1)(x-1)`
`x=3 ; I=1/2Ln8`
`x=2 ; I=1/2Ln3`
`I=1/2(ln8-ln3)=1/2ln(8/3)`