How to integrate x^4/(x^2-1)?

To integrate x^4/(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^4/(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^4 = A(x-1) + B(x+1)

Substitute x=1 to eliminate B and solve for A:

1 = 2A
A = 1/2

Substitute x=-1 to eliminate A and solve for B:

1 = -2B
B = -1/2

Now, rewrite the original integral using the partial fractions:

∫x^4/(x^2-1) dx = ∫1/2(x+1)/(x+1) dx - ∫1/2(x-1)/(x-1) dx

Simplify and integrate:

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

Therefore, the final answer is:

∫x^4/(x^2-1) dx = 1/2(x - 1/2 ln|x^2-1|) + C