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

The integral of x/(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/(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 = A(x^2-1) + B(x^2+1)

We can solve for A and B by substituting in values for x. Letting x = 1, we get:

1 = A(0) + B(2)

So B = 1/2. Similarly, letting x = -1, we get:

-1 = A(0) + B(0)

So A = -1/2. Therefore, we can write:

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

Now we can integrate each term separately. For the first term, we can use the substitution u = x^2+1, du = 2x dx:

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

For the second term, we can use the substitution u = x^2-1, du = 2x dx:

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

Putting it all together, we get:

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