Home > Calculus > Integrals of Rational Functions

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

Answer 1

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

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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Samantha Adkison

To integrate ( \frac{x^3}{4-x^2} ), perform partial fraction decomposition. First, rewrite the expression as ( \frac{x^3}{(2+x)(2-x)} ). Then, decompose it into partial fractions: ( \frac{A}{2+x} + \frac{B}{2-x} ). Solve for (A) and (B), integrate each term separately, and then combine them to get the final result.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.