How to integrate (x^2+1)/(x^2-1)?

To integrate (x^2+1)/(x^2-1), use partial fractions and substitution.

First, factor the denominator: x^2-1 = (x+1)(x-1)

Then, write the fraction as a sum of partial fractions: (x^2+1)/(x^2-1) = A/(x+1) + B/(x-1)

To find A and B, multiply both sides by the common denominator (x+1)(x-1) and simplify: x^2+1 = A(x-1) + B(x+1)

Substitute x=1 to eliminate A: 2 = 2B, so B=1

Substitute x=-1 to eliminate B: 2 = -2A, so A=-1

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

Integrate each term separately using substitution:

∫-1/(x+1) dx = -ln|x+1| + C

∫1/(x-1) dx = ln|x-1| + C

Therefore, the final answer is:

∫(x^2+1)/(x^2-1) dx = -ln|x+1| + ln|x-1| + C

Simplify using logarithmic rules:

∫(x^2+1)/(x^2-1) dx = ln|(x-1)/(x+1)| + C