What's the integral of x^2/(1+x^4)?

The integral of x^2/(1+x^4) can be found using partial fractions and substitution.

To begin, we can use partial fractions to break down the integrand into simpler terms. Let's write:

x^2/(1+x^4) = A/(x^2+1) + B/(x^2-1)

Multiplying both sides by the denominator (1+x^4) and simplifying, we get:

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

Substituting x=1 and x=-1, we can solve for A and B:

A = 1/2, B = -1/2

Therefore, we can rewrite the integrand as:

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

Using substitution, we can now integrate each term separately:

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

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

Therefore, the final answer is:

∫x^2/(1+x^4) dx = 1/2x - 1/2x + C = C