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

1 Answer
Apr 11, 2017

#-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#