What's the integral of x^3/(x^2-1)?

The integral of x^3/(x^2-1) is (1/2)ln|x^2-1| - (1/2)x - (1/2)ln|x+1| + C.

To solve this integral, we can use partial fractions. First, we factor the denominator as (x+1)(x-1). Then, we write the fraction as:

x^3/(x^2-1) = A/(x+1) + B/(x-1)

Multiplying both sides by (x+1)(x-1), we get:

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

Setting x=1, we get A=1/2. Setting x=-1, we get B=-1/2. Therefore, we can write the original fraction as:

x^3/(x^2-1) = (1/2)/(x+1) - (1/2)/(x-1)

Now we can integrate each term separately:

∫(1/2)/(x+1) dx = (1/2)ln|x+1| + C1

∫-(1/2)/(x-1) dx = -(1/2)ln|x-1| + C2

Combining the two integrals and simplifying, we get:

∫x^3/(x^2-1) dx = (1/2)ln|x^2-1| - (1/2)x - (1/2)ln|x+1| + C

Therefore, the integral of x^3/(x^2-1) is (1/2)ln|x^2-1| - (1/2)x - (1/2)ln|x+1| + C.