Integration Using Partial Fractions

Integration by partial fractions is a technique used to integrate rational functions, meaning functions of the form

$$\frac{f(x)}{g(x)}$$​

where f(x) and g(x$)$ are polynomials, and the degree of the numerator is less than the degree of the denominator.

If the numerator’s degree is greater or equal, you first do polynomial long division and then apply partial fractions.

Why Use Partial Fractions?

Some rational functions look complicated but can be broken into simpler fractions that are easy to integrate.

Example:

$$\frac{1}{x^{2}}=\frac{1}{(x+1)(x-1)}$$​

This can be decomposed into:

$$\frac{1}{x^{2}-1}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)$$

Now, each piece is easy to integrate.

Steps for Partial Fraction Decomposition

Step 1: Factor the denominator

Find all linear or irreducible quadratic factors.

Step 2: Set up the partial fractions

The setup depends on factor type.

Types of Denominators and Their Forms

A. Distinct Linear Factors

example:

$$\frac{f(x)}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}$$​

B. Repeated Linear Factors

example:

$$\frac{A}{x+2}+\frac{B}{(x+2)^{2}}+\frac{C}{(x+2)^{3}}$$​

C. Irreducible Quadratic Factors

example:

​$$\frac{Ax+B}{x^{2}+4}$$

D. Repeated Quadratic Factors

example:

$$\frac{Ax+B}{x^{2}+1}+\frac{Cx+B}{(x^{2}+1)^{2}}$$

Example 1: Distinct Linear Factors

$$dx\$\$$$

denominator factor:

$$x^{2}-4 = (x+2)(x-2)$$

Set up:

$$\frac{5}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}$$ multiply $$(x-2)(x+2)$$​ both side then

$$5=(x+2)A+(x-2)B$$ and substitution $$x=2 and x=-2$$ then

$$5 = 4A$$ therefore $$A=\frac{5}{4}$$

$$5= -4B$$ therefore $$B=-\frac{5}{4}$$​

Integrate:

$$\frac{5}{(x-2)(x+2)} = \frac{5}{4}\frac{1}{x-2} -\frac{5}{4}\frac{1}{x+2}$$

$$=\frac{5}{4}\ln|x-2|-\frac{5}{4}\ln|x+2|+C$$

Example 2: Repeated Linear Factor

$$\int\frac{2}{x(x-1)^2}dx$$

Set up:

$$\frac{2}{x(x+1)^{2}}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}$$​ multiply $$x(x+1)^{2}$$​ both side then

$$2=(x-1)^{2}A+x(x-1)B+xC$$ and substitution $$x=-1,0,1$$

$$x=0$$,$$2=A+0\cdot(-1)\cdot(B)+0\cdot(C)$$ therefore $$A=2$$

$$x=1$$,$$2=0\cdot(A)+1\cdot0\cdot(B)+C$$ therefore $$C=2$$

$$x=-1$$,$$2=4\cdot2+2\cdot(B)+2$$ therefore $$B=-4$$

Integrate:

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

$$=2\ln|x|-4\ln|x-1|-\frac{2}{x-1}+C$$

Key Takeaways

✔ Use partial fractions only when integrating rational functions.
✔ Factor the denominator first.
✔ Choose the correct decomposition form based on factor type.
✔ Solve for coefficients using substitution or comparing coefficients.
✔ Integrate each simple fraction individually.