How do you integrate #x^3 / (4-x^2)#?
1 Answer
Apr 11, 2017
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
Then:
The first integral is simple:
#=-1/2x^2+4intx/(4-x^2)dx#
For the second integral, try the substitution
#=-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#