Question

How do you integrate `x^3 / (4-x^2)`?

Answer 1

`-1/2x^2-2lnabs(4-x^2)+C`

Explanation:

We can rewrite the original function:

`x^3/(4-x^2)=(-x(4-x^2)+4x)/(4-x^2)`

`color(white)(x^3/(4-x^2))=-x+(4x)/(4-x^2)`

If you're uncomfortable with this method of simplification, you can also perform the long division for `x^3/(4-x^2)` to find that these are equivalent expressions.

Then:

`intx^3/(4-x^2)dx=-intxdx+4intx/(4-x^2)dx`

The first integral is simple:

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

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

`=-1/2x^2-2int(-2x)/(4-x^2)dx`

`=-1/2x^2-2int1/udu`

This is a common integral:

`=-1/2x^2-2lnabsu`

`=-1/2x^2-2lnabs(4-x^2)+C`