Partial Fraction Technique

Welcome to this guide on integration using Partial Fraction Decomposition on the FiveHive Calculus BC course. This article will guide you through how to perform the partial fraction integration technique. Partial fraction is a technique that focuses on splitting a fraction into a sum of multiple fractions; the example below demonstrate this technique well.

Consider this integral:

$\int \frac{x+1}{x^2-5x+6}dx$

This integral looks scary, but notice that we can factor the denominator: $x^2-5x+6=(x-2)(x-3)$. Let's first take at look at fraction addition:

$\frac{A}{B}+\frac{C}{D}=\frac{AD+BC}{BD}$

Since the integrand is a fraction, and we successfully write the denominator as a product, we should be able to split the fraction into a sum of two fraction.

Assume we have split the fraction like this:

$\frac{x+1}{(x-2)(x-3)}=\frac{A}{(x-2)}+\frac{B}{(x-3)}$

Here A and B are different constants, let's try to combine them together:

$\frac{A(x-3)+B(x-2)}{(x-2)(x-3)}=\frac{Ax-3A+Bx-2B}{(x-2)(x-3)}=\frac{(A+B)x+(-3A-2B)}{(x-2)(x-3)}$

Notice that we reformed the denominator into $\text{constant}\cdot x+\text{constant}$ in the original integrand, which does indeed look like $x+1$ in the original integrand, thus we have this relation:

$\{\begin{array}{l}A+B=1\\ -3A-2B=1\end{array}$

This linear set of equation can be easily solved, which gives the solution of $A=-3$ and $B=4$. Thus we turned this integrand into a fraction of sum:

$\int \frac{x+1}{x^2-5x+6}dx=\int \left(-\frac{3}{(x-2)}+\frac{4}{(x-3)}\right)dx$

Then we can evaluate the integral easily:

$\int \left(-\frac{3}{(x-2)}+\frac{4}{(x-3)}\right)dx=-3\ln |x-2|+4\ln |x-3|+C$

In general, partial fraction follow this process:

Consider a integrand where the denominator can be factored, first split the fraction into a sum of fractions:

$\frac{Ax+B}{(x+C)(x+D)}=\frac{E}{(x+C)}+\frac{F}{(x+D)}=\frac{(E+F)x+(DE+CF)}{(x+C)(x+D)}$

Where $A,B,C,D,E$, and $F$ are all constants, thus:

$\{\begin{array}{l}E+F=A\\ DE+CF=B\end{array}$

Then solve for $E$ and $F$ to complete the fraction split.

Here is another example:

$\int \frac{2x-1}{x^2-4x+3}dx=\int \frac{x+1}{(x-3)(x-1)}dx$

Split the fraction into a sum of fraction:

$\frac{A}{x-3}+\frac{B}{x-1}=\frac{A(x-1)+B(x-3)}{(x-3)(x-1)}=\frac{(A+B)x+(-A-3B)}{(x-3)(x-1)}$

Solve for $A$ and $B$:

$\{\begin{array}{l}A+B=1\\ -A-3B=1\end{array}$

Thus $A=2$ and $B=-1$, and the integral turns to

$\int \left(\frac{2}{x-3}-\frac{1}{x-1}\right)dx=2\ln |x-3|-\ln |x-1|+C$

Practice

Evaluate the following integrals