∫ dx/x²-1

In the above equation before - one is x square.

1/(x^2 -1)=1/(x+1)(x-1)=(1/2)((1/(x+1))-(1/(x-1)))

Answer=(1/2)ln|(x+1)/(x-1)|

1/(x^2 -1)=1/(x+1)(x-1)=(1/2)((1/(x-1))-(1/(x+1)))

Answer=(1/2)ln|(x-1)/(x+1)|

this one is the right answer dude.....well all the best...:)

integration of 1/(1-x²) is tan^-1 x.

Thus here in this problem, if we take a minus sign common,then we will get the equation in this form. thus the answer is

-tan^-1x