TBIL-Cal Partial Fractions (TI6)

Skip to main content

Section 5.6 Partial Fractions (TI6)

Learning Outcomes

I can integrate functions using the method of partial fractions.

Subsection 5.6.1 Activities

Activity 5.6.1.

Consider

\int \frac{x^{2} + x + 1}{x^{3} + x} d x .

Which substitution would you choose to evaluate this integral?

$u = x^{3}$
$u = x^{3} + x$
$u = x^{2} + x + 1$
Substitution is not effective

Activity 5.6.2.

Using the method of substitution, which of these is equal to

\int \frac{5}{x + 7} d x ?

$5 \ln | x + 7 | + C$
$\frac{5}{7} \ln | x + 7 | + C$
$5 \ln | x | + 5 \ln | 7 | + C$
$\frac{5}{7} \ln | x | + C$

Observation 5.6.3.

To avoid repetitive substitution, the following integral formulas will be useful.

\int \frac{1}{x + b} d x = \ln | x + b | + C

\int \frac{1}{(x + b)^{2}} d x = - \frac{1}{x + b} + C

\int \frac{1}{x^{2} + b^{2}} d x = \frac{1}{b} \arctan (\frac{x}{b}) + C

Activity 5.6.4.

Which of the following is equal to

\frac{1}{x} + \frac{1}{x^{2} + 1} ?

$\frac{2 x}{x^{2} + x + 1}$
$\frac{x^{3} + x}{x^{2} + x + 1}$
$\frac{2 x}{x^{3} + x}$
$\frac{x^{2} + x + 1}{x^{3} + x}$

Activity 5.6.5.

Based on the previous activities, which of these is equal to

\int \frac{x^{2} + x + 1}{x^{3} + x} d x ?

$\ln | x | + \arctan (x) + C$
$\ln | x^{2} + x + 1 | + C$
$\ln | x^{3} + x | + C$
$\arctan (x^{3} + x) + C$

Activity 5.6.6.

Suppose we know

\frac{10 x - 11}{x^{2} + x - 2} = \frac{7}{x - 1} + \frac{3}{x + 2} .

Which of these is equal to

\int \frac{10 x - 11}{x^{2} + x - 2} d x ?

$7 \ln | x - 1 | + 3 \arctan (x + 2) + C$
$7 \ln | x - 1 | + 3 \ln | x + 2 | + C$
$7 \arctan (x - 1) + 3 \arctan (x + 2) + C$
$7 \arctan (x - 1) + 3 \ln | x + 2 | + C$

Observation 5.6.7.

To find integrals like

\int \frac{x^{2} + x + 1}{x^{3} + x} d x

and

\int \frac{10 x - 11}{x^{2} + x - 2} d x,

we’d like to decompose the fractions into simpler partial fractions that may be integrated with these formulas

\int \frac{1}{x + b} d x = \ln | x + b | + C

\int \frac{1}{(x + b)^{2}} d x = - \frac{1}{x + b} + C

\int \frac{1}{x^{2} + b^{2}} d x = \frac{1}{b} \arctan (\frac{x}{b}) + C

Fact 5.6.8. Partial Fraction Decomposition.

Let

\frac{p (x)}{q (x)}

be a rational function, where the degree of

p

is less than the degree of

q .

Linear Terms: Let $(x - a)^{n}$ divide $q (x) .$ Then the decomposition of $\frac{p (x)}{q (x)}$ will contain the terms

$\frac{A_{1}}{(x - a)} + \frac{A_{2}}{(x - a)^{2}} + \dots + \frac{A_{n}}{(x - a)^{n}} .$
Quadratic Terms: Let $(x^{2} + b x + c)^{n}$ divide $q (x),$ where $x^{2} + b x + c$ is irreducible. Then the decomposition of $\frac{p (x)}{q (x)}$ will contain the terms

$\frac{B_{1} x + C_{1}}{x^{2} + b x + c} + \frac{B_{2} x + C_{2}}{(x^{2} + b x + c)^{2}} + \dots + \frac{B_{n} x + C_{n}}{(x^{2} + b x + c)^{n}} .$

Example 5.6.9.

Following is an example of a rather involved partial fraction decomposition.

\begin{aligned} \frac{7 x^{6} - 4 x^{5} + 41 x^{4} - 20 x^{3} + 24 x^{2} + 11 x + 16}{x (x - 1)^{2} (x^{2} + 4)^{2}} \\ = & \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^{2}} + \frac{D x + E}{x^{2} + 4} + \frac{F x + G}{(x^{2} + 4)^{2}} \end{aligned}

Using some algebra, it’s possible to find values for

A

through

G

to determine

\begin{aligned} \frac{7 x^{6} - 4 x^{5} + 41 x^{4} - 20 x^{3} + 24 x^{2} + 11 x + 16}{x (x - 1)^{2} (x^{2} + 4)^{2}} \\ = & \frac{1}{x} + \frac{2}{x - 1} + \frac{3}{(x - 1)^{2}} + \frac{4 x + 5}{x^{2} + 4} + \frac{6 x + 7}{(x^{2} + 4)^{2}} . \end{aligned}

Activity 5.6.10.

Which of the following is the form of the partial fraction decomposition of

\frac{x^{3} - 7 x^{2} - 7 x + 15}{x^{3} (x + 5)} ?

$\frac{A}{x} + \frac{B}{x + 5}$
$\frac{A}{x^{3}} + \frac{B}{x + 5}$
$\frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x^{3}} + \frac{D}{x + 5}$
$\frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x^{3}} + \frac{D x + E}{x + 5}$

Activity 5.6.11.

Which of the following is the form of the partial fraction decomposition of

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

$\frac{A}{x - 3} + \frac{B}{(x - 3)^{2}} + \frac{C}{x^{2} + 4} + \frac{D}{(x^{2} + 4)^{2}}$
$\frac{A}{x - 3} + \frac{B}{(x - 3)^{2}} + \frac{C x + D}{(x^{2} + 4)^{2}}$
$\frac{A}{x - 3} + \frac{B}{(x - 3)^{2}} + \frac{C}{x^{2} + 4} + \frac{D x + E}{(x^{2} + 4)^{2}}$
$\frac{A}{x - 3} + \frac{B}{(x - 3)^{2}} + \frac{C x + D}{x^{2} + 4} + \frac{E x + F}{(x^{2} + 4)^{2}}$

Activity 5.6.12.

Consider that the partial decomposition of

\frac{x^{2} + 5 x + 3}{(x + 1)^{2} x}

is

\frac{x^{2} + 5 x + 3}{(x + 1)^{2} x} = \frac{A}{x + 1} + \frac{B}{(x + 1)^{2}} + \frac{C}{x} .

What equality do we obtain if we multiply both sides of the above equation by

(x + 1)^{2} x ?

$x^{2} + 5 x + 3 = A x (x + 1) + B x + C (x + 1)^{2}$
$x^{2} + 5 x + 3 = A (x + 1) + B (x + 1)^{2} + C x$
$x^{2} + 5 x + 3 = A x (x + 1) + B x + C (x + 1)$
$x^{2} + 5 x + 3 = A x (x + 1) + B x^{2} + C (x + 1)^{2}$

Activity 5.6.13.

Use your choice in Activity 5.6.12 (which must hold for any

x

value) to answer the following.

(a)

By substituting

x = 0

into the equation, we may find:

$A = 1$
$B = - 2$
$C = 3$

(b)

By substituting

x = - 1

into the equation, we may find:

$A = - 4$
$B = 1$
$C = 5$

Activity 5.6.14.

Using the results of Activity 5.6.13, show how to rewrite our choice from Activity 5.6.12

? x^{2} + ? x = A x^{2} + A x .

What value of

A

satisfies this equation?

$- 2$
$3$
$4$
$- 5$

Activity 5.6.15.

By using the form of the decomposition

\frac{x^{2} + 5 x + 3}{(x + 1)^{2} x} = \frac{A}{x + 1} + \frac{B}{(x + 1)^{2}} + \frac{C}{x}

and the coefficients found in Activity 5.6.13 and Activity 5.6.14, evaluate

\int \frac{x^{2} + 5 x + 3}{(x + 1)^{2} x} d x .

Activity 5.6.16.

Given that

\frac{x^{3} - 7 x^{2} - 7 x + 15}{x^{3} (x + 5)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x^{3}} + \frac{D}{x + 5}

do the following to find

A, B, C,

and

D .

(a)

Eliminate the fractions to obtain

x^{3} - 7 x^{2} - 7 x + 15 = A (?) (?) + B (?) (?) + C (?) + D (?) .

(b)

Plug in an

x

value that lets you find the value of

C .

(c)

Plug in an

x

value that lets you find the value of

D .

(d)

Use other algebra techniques to find the values of

A

and

B .

Activity 5.6.17.

Given your choice in Activity 5.6.16 Find

\int \frac{x^{3} - 7 x^{2} - 7 x + 15}{x^{3} (x + 5)} d x .

Activity 5.6.18.

Consider the rational expression

\frac{2 x^{3} + 2 x + 4}{x^{4} + 2 x^{3} + 4 x^{2}} .

Which of the following is the partial fraction decomposition of this rational expression?

$\frac{1}{x} + \frac{1}{x^{2}} + \frac{2 x - 1}{x^{2} + 2 x + 4}$
$\frac{2}{x} + \frac{0}{x^{2}} + \frac{- 1}{x^{2} + 2 x + 4}$
$\frac{0}{x} + \frac{1}{x^{2}} + \frac{- 1}{x^{2} + 2 x + 4}$
$\frac{0}{x} + \frac{1}{x^{2}} + \frac{2 x - 1}{x^{2} + 2 x + 4}$

Activity 5.6.19.

Given your choice in Activity 5.6.18 Find

\int \frac{2 x^{3} + 2 x + 4}{x^{4} + 2 x^{3} + 4 x^{2}} d x .

Activity 5.6.20.

Given that

\frac{2 x + 5}{x^{2} + 3 x + 2} = \frac{- 1}{x + 2} + \frac{3}{x + 1},

find

\int_{0}^{3} \frac{2 x + 5}{x^{2} + 3 x + 2} d x .

Activity 5.6.21.

Evaluate

\int \frac{4 x^{2} - 3 x + 1}{(2 x + 1) (x + 2) (x - 3)} d x .

Subsection 5.6.2 Videos

Figure 111. Video: I can integrate functions using the method of partial fractions

Subsection 5.6.3 Exercises

Exercises available at https://tbil.org/calculus/preview/exercises/#/bank/TI6/