Integration by Partial Fraction

Prerequisites

• Basic Integration
• Basic Algebra
• du substitution

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Method

Partial Fraction is thinking one fraction as a sums of multiple fractions.
(Especially the ones with factorable denominator)

\[\begin{gather} \frac{f(x)}{g(x)h(x)} = \frac{a}{g(x)} + \frac{b}{h(x)} \\\\ \implies f(x) = a \cdot h(x) + b \cdot g(x) \end{gather}\]

So it becomes easier to integrate with sums because we can now distribute integral.

\[\begin{gather} \int \frac{f(x)}{g(x)h(x)} \,dx = \int \frac{a}{g(x)} + \frac{b}{h(x)} \,dx \end{gather}\]

Plug in values to isolate the variable

\[\begin{gather} a \cdot g(x) + b \cdot h(x) = f(x) \end{gather}\]

We can let $x = k$ which makes $g(k) = 0$.

So the equation becomes

\[\begin{gather} a \cdot g(k) + b \cdot h(k) = f(k) \\ \implies a \cdot 0 + b \cdot h(k) = f(k) \end{gather}\]

Then we can solve for b since $h(k)$ and $f(k)$ is now evaluated to constant.

We can repeat this to solve a or other variables.

Systems of equation

We can also use systems of equation to solve for a and b.

\[\begin{gather} \text{let } g(x) = x+2 \,, h(x) = x+3 \,, f(x) = 2x+5\\ a \cdot g(x) + b \cdot h(x) = f(x) \\ = a (x+2) + b (x+3) = 2x+5 \\ = ax + 2a + bx + 3b =2x + 5 \\ = (a+b)x + 2a + 3b = 2x + 5 \\ \implies a+b = 2 \\ \implies 2a + 3b = 5 \end{gather}\]

Then we can solve for a and b.

\[\begin{gather} a+b = 2 \\ 2a + 3b = 5 \\\\ 2a + 3b = 5 \\ 2a + 2b = 4 \\\\ 2a - 2a + 3b - 2b = 1 \\ b = 1 \,, a=1 \end{gather}\]

Integration Example

\[\begin{align} \int \frac{x}{x^2 -4} \,dx &= \int \frac{x}{(x+2)(x-2)} \,dx \\ &= \int \frac{a}{x+2} + \frac{b}{x-2} \,dx \\ \because x &= a(x-2) + b(x+2) \end{align}\]

Then plug in values to isolate the variable.

\[\begin{gather} \text{let} \, x=2 \\ x = a(x-2) + b(x+2) \\ 2 = a(2-2) + b(2+2) \\ 2 = a \cdot 0 + b\cdot 4\\ 2= 4b\\ \therefore b=\frac{1}{2}\\ \text{let} \, x= -2 \\ x = a(x-2) + b(x+2) \\ -2 = a(-2-2) + b(-2+2) \\ 2 = a \cdot (-4) + b\cdot 0\\ -2= -4a\\ \therefore a=\frac{1}{2} \end{gather}\]

Plug a and b into original equation,

\[\begin{align} \int \frac{a}{x+2} + \frac{b}{x-2} \,dx &= \int \frac{1/2}{x+2} + \frac{1/2}{x-2} \,dx \\ &= \frac{1}{2} \int \frac{1}{x+2} + \frac{1}{x-2} \,dx \\ &= \frac{1}{2} \left( \ln|x+2|+ \ln|x-2| \right) + C\\ &= \frac{1}{2} \ln|(x+2)(x-2)| + C\\ &= \frac{1}{2} \ln|x^2 -4| + C \end{align}\]

We can also use du substitution to check our work in this example.