Question

Integration of x³/4-x²

Answer 1

We can rewrite the original function:

(x³/(4 - x²) = (-x(4 - x²) + 4x)/(4 - x²)

(x³/(4 - x²) = -x + (4x)/(4 - x²)

If you're uncomfortable with this method of simplification, you can also perform the long division for x³/(4-x²) to find that these are equivalent expressions.

intx³/(4 - x²)dx = -intxdx + 4intx/(4 - x²)dx

The first integral is simple:

=-1/2x² + 4intx/(4-x²)dx

For the second integral, try the substitution u = 4 − x² .This implies that du = -2xdx, We have the derivative off by a factor of −2 already in the numerator:

=-1/2x² - 2int(-2x)/(4 - x²)dx

=-1/2x² - 2int1/udu

This is a common integral:

=-1/2x² - 2lnabsu

=-1/2x² - 2lnabs(4 - x²) + C