# What is `int 3/(x^2-1) dx`

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The other way to do the partial fractions is

`A/(x-1)+B/(x+1)=3/((x+1)(x-1))` multiplying by `(x+1)(x-1)` we get

`A(x+1)+B(x-1)=3`

Set `x=1` to get

` A(2)+B(0)=3`

and this gives `A=3/2`

Setting `x=-1` we get

`A(0)+B(-2)=3`

`B = -3/2`

This is much easier than solving the system of equations...

The integral `int 3/(x^2 - 1) dx` has to be determined.

`3/(x^2 - 1)`

=> `3/((x - 1)(x + 1))`

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

=> `(Ax + A + Bx - B)/((x+1)(x-1))`

This gives Ax + A + Bx - B = 3

=> A + B = 0 and A - B = 3

=> A = `3/2` and B = `-3/2`

The integral is now: `int 3/(2*(x - 1)) - 3/(2*(x+1)) dx`

=> `(3/2)*ln (x - 1) - (3/2)*ln(x + 1)`

=> `(3/2)*ln((x - 1)/(x + 1))`

**The required integral is `(3/2)*ln((x - 1)/(x + 1)) + C` **