Integration with Partial Fractions Examples 1

# Integration with Partial Fractions Examples 1

Recall the material outlined on the Integration with Partial Fractions. We will now look at some examples of integrating functions through partial fractions. More examples can be found on the Integration with Partial Fractions Examples 2 page.

## Example 1

Integrate $f(x) = \frac{2x^2 + 3}{x^3 - 2x^2 + x}$.

We start with a proper rational function, so we do not need to do long division. We will now factor the denominator to obtain that $x^3 -2x^2 + x = x(x-1)^2$. Hence it follows that since one of the linear factors is repeated that for some A, B, and C:

(1)
\begin{align} \frac{2x^2 + 3}{x^3 - 2x^2 + x} = \frac{A}{x} + \frac{B}{(x - 1)} + \frac{C}{(x - 1)^2} \\ \frac{2x^2 + 3}{x^3 - 2x^2 + x} = \frac{A(x-1)^2}{x} + \frac{Bx(x-1)}{(x - 1)} + \frac{Cx}{(x - 1)^2} \\ \frac{2x^2 + 3}{x^3 - 2x^2 + x} = \frac{A(x-1)^2 + Bx(x-1) + Cx}{x(x-1)^2} \end{align}

Hence it follows that $2x^2 + 3 = A(x-1)^2 + Bx(x-1) + Cx$. And now we can solve for A, B, and C by choosing values of x. First let's let x = 1. Then it follows that 5 = C. Now let's choose x = 0, and it follows that A = 3. Lastly let's choose x = 2, and get that 11 = A + 2B + 2C. We know the values of A and C though, so we find that B = -1 by substitution. Hence we get that:

(2)
\begin{align} \frac{2x^2 + 3}{x^3 - 2x^2 + x} = \frac{3}{x} + \frac{-1}{(x - 1)} + \frac{5}{(x - 1)^2} \\ \end{align}

Now we can integrate:

(3)
\begin{align} \int f(x) \: dx = \int \frac{3}{x} + \frac{-1}{(x - 1)} + \frac{5}{(x - 1)^2} \: dx \\ \quad \int f(x) \: dx = 3 \ln | x | - \ln | x - 1 | - \frac{5}{x - 1} + C \end{align}

## Example 2

Integrate $f(x) = \frac{x^2 + 5x + 2}{(x + 1)(x^2 + 1)}$.

We already have the denominator factored, however, this time we have an irreducible quadratic factor in the denominator. Hence we know that:

(4)
\begin{align} \frac{x^2 + 5x + 2}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1} \\ \frac{x^2 + 5x + 2}{(x + 1)(x^2 + 1)} = \frac{A(x^2 + 1) + (Bx + C)(x + 1)}{(x +1)(x^2 + 1)} \\ \end{align}

So it follows that $x^2 + 5x + 2 = A(x^2 + 1) + (Bx + C)(x+1)$. Let's let x = -1, it thus follows that 2A = -2, or more appropriately A = -1. When we let x = 0, we get that A + C = 2, but we know that A = -1, so C = 3. When we let x = 1, then 2A + 2B + 2C = 8, and by substitution B = 2. Hence it follows that:

(5)
\begin{align} f(x) = \frac{-1}{x + 1} + \frac{2x + 3}{x^2 + 1} \\ \int f(x) \: dx = \int \frac{-1}{x + 1} + \frac{2x + 3}{x^2 + 1} \: dx \\ \int f(x) \: dx = - \ln | x + 1 | + \int \frac{2x + 3}{x^2 + 1} \: dx \\ \quad \int f(x) \: dx = - \ln | x + 1 | + \int \frac{2x}{x^2 + 1} \: dx + \int \frac{3}{x^2 + 1} \: dx\\ \end{align}

We will omit fully completing this example, but integration by partial fractions can be repeated for this example.

## Example 3

Evaluate the following integral: $\int \frac{x - 9}{(x + 5)(x - 2)} \: dx$.

We first recognize that for some A and B:

(6)
\begin{align} \frac{x - 9}{(x + 5)(x - 2)} = \frac{A}{(x + 5)} + \frac{B}{(x - 2)} \\ \frac{x - 9}{(x + 5)(x - 2)} = \frac{A(x-2) + B(x+5)}{(x+5)(x-2)} \end{align}

Hence $x- 9 = A(x -2) + B(x + 5)$. When we let x = -5, then -7A = -14, or rather, A = 2. When we let x = 2, we get that B = -1. Hence we get that:

(7)
\begin{align} \frac{x - 9}{(x + 5)(x - 2)} = \frac{2}{(x + 5)} - \frac{1}{(x - 2)} \\ \quad \int \frac{2}{(x + 5)} - \frac{1}{(x - 2)} \: dx = 2 \ln | x + 5 | - \ln | x - 2 | + C \end{align}

## Example 4

Evaluate the following integral: $\int \frac{2}{2x^2 + 3x + 1} \: dx$.

Let's first factor the denominator to get that $2x^2 +3x + 1 = (2x + 1)(x + 1)$. He thus know that for some A and B:

(8)
\begin{align} \frac{2}{(2x + 1)(x + 1)} = \frac{A}{(2x + 1)} + \frac{B}{(x + 1)} \\ \frac{2}{(2x + 1)(x + 1)} = \frac{A(x +1) + B(2x + 1)}{(2x + 1)(x + 1)} \end{align}

So then $2 = A(x+1) + B(2x + 1)$. If we let x = -1, then -B = 2, so B = -2. If we let x = -1/2, then we get that A/2 = 2, or rather A = 4. Hence we can now integrate the function:

(9)
\begin{align} \int \frac{4}{(2x + 1)} - \frac{2}{(x + 1)} \: dx \\ \quad 2 \ln | 2x + 1 | - 2 \ln | x + 1 | + C \end{align}

## Example 5

Evaluate the following integral: $\int \frac{x^2 + 1}{(x-3)(x-2)^2} \: dx$.

For some A, B, and C, we get that:

(10)
\begin{align} \frac{x^2 + 1}{(x-3)(x-2)^2} = \frac{A}{(x-3)} + \frac{B}{(x - 2)} + \frac{C}{(x - 2)^2} \\ \frac{x^2 + 1}{(x-3)(x-2)^2} = \frac{A(x-2)^2 + B(x-3)(x-2) + C(x-3)}{(x-3)(x-2)^2} \\ \end{align}

It thus follows that $x^2 + 1 = A(x - 2)^2 + B(x - 3)(x - 2) + C(x - 3)$. If we let x = 3, we get that A = 10. If we let x = 2, we get that -C = 5, or more appropriately that C = -5. If we let x = 1, we get that A + 2B - 2C = 2. By substitution we get B = -9, and now we can integrate:

(11)
\begin{align} \frac{x^2 + 1}{(x-3)(x-2)^2} = \frac{10}{(x-3)} - \frac{9}{(x - 2)} - \frac{5}{(x - 2)^2} \\ \int \frac{10}{(x-3)} - \frac{9}{(x - 2)} - \frac{5}{(x - 2)^2} \: dx = 10 \ln | x - 3 | - 9 \ln | x - 2 | + \frac{5}{(x - 2)} + C \end{align}

## Example 6

Evaluate the following integral: $\int \frac{x^3 + x^2 + 2x + 1}{(x^2 + 1)(x^2 + 2)} \: dx$.

In this example, the denominator consists of two irreducible quadratic factors. Hence we know that for some A, B, C, and D:

(12)
\begin{align} \frac{x^3 + x^2 + 2x + 1}{(x^2 + 1)(x^2 + 2)} = \frac{Ax + B}{(x^2 + 1)} + \frac{Cx + D}{(x^2 + 2)} \\ \frac{x^3 + x^2 + 2x + 1}{(x^2 + 1)(x^2 + 2)} = \frac{(Ax + B)(x^2 + 2) + (Cx + D)(x^2 + 1)}{(x^2 + 1)(x^2 + 2)} \end{align}

Hence we know that $x^3 + x^2 + 2x + 1 = (Ax + B)(x^2 + 2) + (Cx + D)(x^2 +1)$.

First let's let x = 0. Hence we get that $1 = 2B + D$. Suppose B = 1, and D = -1. Continuing forward, let x = 1. We get that $5 = (A + B)(3) + (C + D)(2)$, or rather $5 = 3A + 3B + 2C + 2D$. Letting B = 1 and D = -1, we get that $5 = 3A + 3 + 2C - 2$, or more appropriately $4 = 3A + 2C$. Let's let A = 2, then C = -1. Hence we have a combination:

(13)
\begin{align} A = 2 \\ B = 1 \\ C = -1 \\ D = -1 \end{align}

Hence we can now integrate the function:

(14)
\begin{align} \frac{x^3 + x^2 + 2x + 1}{(x^2 + 1)(x^2 + 2)} = \frac{2x + 1}{(x^2 + 1)} + \frac{-x -1}{(x^2 + 2)} \end{align}