# Integration with Partial Fractions

Integration with partial fractions is a useful technique to make a rational function simpler to integrate. Before continuing on to read the rest of this page, you should consult the various wikis related to partial fraction decomposition.

## Examples

The best way to learn this technique of integration is through example.

Find the antiderivative:

\[\int { \frac { 1 }{ x(x+1) } dx } \]

In this form, the integral is not so simple. However, the integrand may be split into two parts using Partial Fraction Decomposition.

\[\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \]

\[1 = A(x+1) + B(x) \]

Letting \( \displaystyle x = -1 \) and \(\displaystyle x = 0 \), we see that \(\displaystyle A = 1 \) and \(B = -1 \). We now have

\[\int { \left( \frac { 1 }{ x } -\frac { 1 }{ x+1 } \right) dx = \int { \frac { dx }{ x } -\int { \frac { dx }{ x+1 } =\ln { \left| x \right| -\ln { \left| x+1 \right| +C } } } } } \]

where \(\displaystyle C \) is the arbitrary constant of integration. \(\displaystyle \square \)

Integrate the following:

\[\int { \frac { 2x+3 }{ { x }^{ 2 }-9 } dx } \]

Once again, this is not a simply integrated function in the form it is in right now. However, it can be made significantly simpler with Partial Fractions.

\[\frac{2x+3}{(x+3)(x-3)} = \frac{A}{x+3} + \frac{B}{x-3} \]

\[2x+3 = A(x-3) + B(x+3) \]

Letting \(\displaystyle x = 3\) and \(\displaystyle x =-3 \), we see that \(\displaystyle A = \frac{1}{2} \) and \(\displaystyle B = \frac{3}{2} \).

\[\int { \left[ \frac { 1 }{ 2 } \left( \frac { 1 }{ x+3 } \right) +\frac { 3 }{ 2 } \left( \frac { 1 }{ x-3 } \right) \right] dx } =\frac { 1 }{ 2 } \int { \frac { dx }{ x+3 } } +\frac { 3 }{ 2 } \int { \frac { dx }{ x-3 } }\]

\[=\frac { 1 }{ 2 } \ln { \left| x+3 \right| +\frac { 3 }{ 2 } \ln { \left| x-3 \right| +C } } \]

where \(\displaystyle C\) is again an arbitrary constant of integration. \(\displaystyle \square\)

## Warning

There are times, however, where it may seem like Partial Fractions is the best way to approach an integral, although another approach would be much simpler in terms of effort and time. Here is such an integral

Find the antiderivative:

\[\int { \frac { 2x+3 }{ { x }^{ 2 }+3x+9 } dx } \]

On first glance, Partial Fractions seems appealing. However, notice that the numerator is simply the derivative of the denominator! This means the integral can be found using u-substitution.

Letting \(\displaystyle u = x^2+3x+9\), we find that \(\displaystyle du = (2x+3)dx \). The integral is now

\[\int { \frac { du }{ u } =\ln { \left| u \right| =\ln { \left| { x }^{ 2 }+3x+9 \right| +C } } } \]

Once again, C is the arbitrary constant of integration. \(\displaystyle \square \)

**Cite as:**Integration with Partial Fractions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/integration-with-partial-fractions/