# 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.

Before taking some examples you should remember some simple things :

\[\int { \frac { f(x) }{ g(x) } dx } \]

- Degree of \(f(x)\) must be less than degree of \(g(x)\) , Otherwise use long division firstly.
-Linear functions be prime : .

\(ax+b\) be prime " Not reducible " .

\(ax^2+bx+c\) , so \(a \neq 0 \)

and if \(b^2 -4ac<0\) it's prime (irreducible) , such as : \(x^2+4\) , \(x^2+x+1\)

Otherwise \(b^2-4ac \geq 0 \) " reducible "

Any polynomial of degree \(\geq 3 \) is not prime " reducible ".

## Examples

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

Write the partial fraction decomposition ( hypothesis ) .

\[f(x)=\frac{x}{x^2-x-2} \]

We can easily get :

\[\frac{x}{(x+1)(x-2)} \]

So , the function may be split into two parts using Partial Fraction Decomposition .

\[\frac{x}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2} \]

\[f(x)=\frac{x+10}{x^3+x} \]

Taking a common factor \(x\) , we get :

\[f(x)=\frac{x+10}{x(x^2+1)} \]

Now , look carefully :

\[\frac{x+10}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+c}{x^2+1} \]

Notice : When we have a square polynomial in denominator we have to use Formula for Linear in numerator .

\[f(x)=\frac{1}{(x^3-1)(x^2-1)^2} \]

\[f(x)=\frac{1}{(x-1)(x^2+x+1)((x-1)(x+1))^2} \]

\[f(x)=\frac{1}{(x-1)^3(x+1)^2(x^2+x+1)} \]

\[= \frac{A}{x-1} + \frac{B}{(x-1)^2}+\frac{C}{(x-1)^3} + \frac{D}{(x+1)}+\frac{E}{(x+1)^2} + \frac{Fx+G}{x^2+x+1} \]

Notice : the last term has linear formula , Why ? because its denominator has square polynomial . So , and the fifth term ? why not ? Because the fifth term is already repeated about the fourth term so we don't consider it as square polynomial , Due we care about the square polynomial without repeat in the denominators .

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/