Integration Using Partial Fractions

Integration using partial fractions

When the integrand is a proper rational function (the degree of the numerator is less than the degree of the denominator), it can be decomposed into partial fractions before integration. This simplifies the integration process.

Forms of partial fractions

Here's a table showing some common forms of rational functions and their partial fraction decompositions:

Rational Function	Partial Fraction Decomposition
$\frac{px + q}{(x - a)(x - b)}$ (where $a \ne b$)	$\frac{A}{x - a} + \frac{B}{x - b}$
$\frac{1}{x^2 - a^2}$	$\frac{A}{x - a} + \frac{B}{x + a}$
$\frac{px + q}{(x - a)^2}$	$\frac{A}{x - a} + \frac{B}{(x - a)^2}$
$\frac{px^2 + qx + r}{(x - a)(x - b)(x - c)}$	$\frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c}$
$\frac{px^2 + qx + r}{(x - a)^2(x - c)}$	$\frac{A}{x - a} + \frac{B}{(x - a)^2} + \frac{C}{x - c}$
$\frac{px^2 + qx + r}{(x - a)(x^2 + bx + c)}$ (where $x^2+bx+c$ is irreducible)	$\frac{A}{x - a} + \frac{Bx + C}{x^2 + bx + c}$

Example 1: Find $\int \frac{x^2}{x^2 - 4x + 3} \, dx$.

First, we perform polynomial long division because the degree of the numerator is equal to the degree of the denominator:

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

Now, decompose the fractional part into partial fractions:

$\frac{4x - 3}{(x - 1)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 3}$

$4x - 3 = A(x - 3) + B(x - 1)$

If $x = 1$, $-1 = -2A \Rightarrow A = \frac{1}{2}$.

If $x = 3$, $9 = 2B \Rightarrow B = \frac{9}{2}$.

So, $\frac{x^2}{x^2 - 4x + 3} = 1 + \frac{1/2}{x - 1} + \frac{9/2}{x - 3}$

$\int \frac{x^2}{x^2 - 4x + 3} \, dx = \int \left(1 + \frac{1/2}{x - 1} + \frac{9/2}{x - 3}\right) \, dx = x + \frac{1}{2}\ln|x - 1| + \frac{9}{2}\ln|x - 3| + C$

Example 2: Determine $\int \frac{1}{x^2 - 9} \, dx$.

$\frac{1}{x^2 - 9} = \frac{1}{(x - 3)(x + 3)} = \frac{A}{x - 3} + \frac{B}{x + 3}$

$1 = A(x + 3) + B(x - 3)$

If $x = 3$, $1 = 6A \Rightarrow A = \frac{1}{6}$.

If $x = -3$, $1 = -6B \Rightarrow B = -\frac{1}{6}$.

$\int \frac{1}{x^2 - 9} \, dx = \int \left(\frac{1/6}{x - 3} - \frac{1/6}{x + 3}\right) \, dx = \frac{1}{6}\ln|x - 3| - \frac{1}{6}\ln|x + 3| + C = \frac{1}{6}\ln\left|\frac{x - 3}{x + 3}\right| + C$

Example 3: Find $\int \frac{5x - 2}{x(x + 3)^2} \, dx$.

$\frac{5x - 2}{x(x + 3)^2} = \frac{A}{x} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}$

$5x - 2 = A(x + 3)^2 + Bx(x + 3) + Cx$

If $x = 0$, $-2 = 9A \Rightarrow A = -\frac{2}{9}$.

If $x = -3$, $-17 = -3C \Rightarrow C = \frac{17}{3}$.

Comparing coefficients of $x^2$: $0 = A + B \Rightarrow B = -A = \frac{2}{9}$.

$\int \frac{5x - 2}{x(x + 3)^2} \, dx = \int \left(-\frac{2/9}{x} + \frac{2/9}{x + 3} + \frac{17/3}{(x + 3)^2}\right) \, dx = -\frac{2}{9}\ln|x| + \frac{2}{9}\ln|x + 3| - \frac{17}{3(x + 3)} + C$

Example 4: Find $\int \frac{x^2 - 2x + 4}{(x + 1)(x - 1)} \, dx$.

First, polynomial long division:

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

Now partial fractions:

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

$-2x+5 = A(x+1) + B(x-1)$

If $x=1$, $3=2A$, so $A=3/2$.

If $x=-1$, $7=-2B$, so $B=-7/2$.

$\int \frac{x^2 - 2x + 4}{(x + 1)(x - 1)} \, dx = \int \left(1 + \frac{3/2}{x - 1} - \frac{7/2}{x + 1}\right) \, dx$

$= x + \frac{3}{2}\ln|x - 1| - \frac{7}{2}\ln|x + 1| + C$

Example 5: Integrate $\int \frac{x^2 + 3}{(x+1)(x-2)(x-3)}\,dx$

We decompose the fraction into partial fractions:

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

$x^2+3 = A(x-2)(x-3) + B(x+1)(x-3) + C(x+1)(x-2)$

If $x=-1$: $4 = A(-3)(-4) \implies 4 = 12A \implies A = \frac{1}{3}$

If $x=2$: $7 = B(3)(-1) \implies 7 = -3B \implies B = -\frac{7}{3}$

If $x=3$: $12 = C(4)(1) \implies 12 = 4C \implies C = 3$

So,

$\int \frac{x^2 + 3}{(x+1)(x-2)(x-3)}\,dx = \int \left(\frac{1/3}{x+1} - \frac{7/3}{x-2} + \frac{3}{x-3}\right)dx$

$= \frac{1}{3}\ln|x+1| - \frac{7}{3}\ln|x-2| + 3\ln|x-3| + C$

Example 6: Integrate $\int \frac{x^3 + 2}{x^2 - 1} \, dx$

This is an improper fraction, so we divide first:

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

Now partial fractions:

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

$x+2 = A(x+1) + B(x-1)$

If $x=1$, $3=2A$, so $A=3/2$.

If $x=-1$, $1=-2B$, so $B=-1/2$.

So,

$\int \frac{x^3+2}{x^2-1}\,dx = \int \left(x + \frac{3/2}{x-1} - \frac{1/2}{x+1}\right)dx$

$= \frac{x^2}{2} + \frac{3}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C$

Example 7: Integrate $\int \frac{x^2+1}{x^3+x}\,dx$

$\int \frac{x^2+1}{x^3+x}\,dx = \int \frac{x^2+1}{x(x^2+1)}\,dx = \int \frac{1}{x}\,dx = \ln|x| + C$