Chapter 7 · Integrals

Exercise 7.5 · Q12: $\int\frac{x^3+x+1}{x^2-1}\,dx$

Exercise 7.5Integration by partial fractions · Section 7.5

The question

Integrate the rational function:

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

Exam answer

Let $I = \displaystyle\int\frac{x^3 + x + 1}{x^2 - 1}\,dx$ ⁠.

The fraction is improper. Dividing $x^3 + x + 1$ by $x^2 - 1$ ⁠:

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

Let

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

\Rightarrow\ 2x + 1 = A(x + 1) + B(x - 1)

Putting $x = 1$ ⁠: $3 = 2A$ ⁠, so $A = \frac32$ ⁠.

Putting $x = -1$ ⁠: $-1 = -2B$ ⁠, so $B = \frac12$ ⁠.

\begin{aligned} \therefore\ I &= \int x\,dx + \frac32\int\frac{dx}{x - 1} \\ &\quad + \frac12\int\frac{dx}{x + 1} \end{aligned}

$\therefore\ I = \dfrac{x^2}{2} + \dfrac32\log|x - 1|$ $+ \dfrac12\log|x + 1| + C$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The numerator has degree 3 and the denominator degree 2, so the fraction is improper. Long division gives a polynomial plus a proper fraction: $x + \frac{2x+1}{x^2-1}$ (Section 7.5). The polynomial is integrated directly; the proper fraction is split into $\frac{A}{x-1} + \frac{B}{x+1}$ ⁠.

Given, to find, and the plan

Given: $f(x) = \frac{x^3+x+1}{x^2-1}$ ⁠, defined for every real $x$ except $x = 1$ and $x = -1$ ⁠.
To find: $\displaystyle\int\frac{x^3+x+1}{x^2-1}\,dx$ ⁠.
Plan: (1) See that the fraction is improper. (2) Divide. (3) Split the proper remainder fraction (Table 7.2, row 1) and find $A$ ⁠, $B$ ⁠. (4) Integrate. (5) Differentiate back.

Basics you need

1Improper fractions are divided first

$\frac{P(x)}{Q(x)} = T(x) + \frac{P_1(x)}{Q(x)}$ ⁠, with $T$ a polynomial and $\frac{P_1}{Q}$ proper (Section 7.5; Example 12 does this). Basics: splitting fractions

2Table 7.2, row 1

For $a \ne b$ ⁠: $\frac{px+q}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$ ⁠. Lesson: partial fractions

3Power rule and the logarithm

$\int x\,dx = \frac{x^2}{2} + C$ (standard formula (i)); $\int\frac{dx}{x+k} = \log|x+k| + C$ (formula (xii), Section 7.2). Basics: powers of x

Solution

The fraction is improper

The numerator $x^3 + x + 1$ has degree 3; the denominator $x^2 - 1$ has degree 2. Since $3 \ge 2$ ⁠, the fraction is improper.

A sum of partial fractions like $\frac{A}{x-1} + \frac{B}{x+1}$ is always proper, so it can never equal an improper fraction. The polynomial part must be removed first.

Section 7.5 (improper rational functions). Lesson: partial fractions
Divide $x^3 + x + 1$ by $x^2 - 1$

The leading terms give the first quotient term $\frac{x^3}{x^2} = x$ ⁠. Subtract $x(x^2 - 1) = x^3 - x$ ⁠:
$(x^3 + x + 1) - (x^3 - x) = 2x + 1.$
The remainder $2x + 1$ has degree 1, less than 2, so the division stops:
$\begin{aligned} x^3 + x + 1 &= x(x^2-1) + (2x+1), \\ \frac{x^3+x+1}{x^2-1} &= x + \frac{2x+1}{x^2-1}. \qquad (1) \end{aligned}$
Check: $x(x^2-1) + 2x + 1$ $= x^3 - x + 2x + 1$ $= x^3 + x + 1$ ⁠. ✓ The quotient $x$ is the polynomial part $T(x)$ ⁠; the remainder fraction is proper.

Long division of polynomials, Section 7.5 (compare Example 12). Basics: splitting fractions
Split the proper fraction

Factor $x^2 - 1 = (x-1)(x+1)$ and write
$\frac{2x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}. \qquad (2)$
Multiply by $(x-1)(x+1)$ ⁠:
$2x + 1 = A(x+1) + B(x-1). \qquad (3)$
Two different linear factors: row 1 of Table 7.2. (3) is a polynomial identity, valid for every real $x$ ⁠.

Table 7.2, row 1; Remark after Example 11, Section 7.5.
Find $A$ and $B$

$x = 1$ ⁠: $3 = 2A$ ⁠, so $A = \frac32$ ⁠.

$x = -1$ ⁠: $-1 = -2B$ ⁠, so $B = \frac12$ ⁠.

Confirm: the right side of (3) is $(A+B)x + (A-B)$ ⁠:
$\begin{aligned} \text{coefficient of } x:&\quad A + B = \tfrac32 + \tfrac12 = 2, \\ \text{constant term}:&\quad A - B = \tfrac32 - \tfrac12 = 1. \end{aligned}$
Both match the left side $2x + 1$ ⁠.

Basics: splitting fractions
Integrate

From (1) and (2):
$\frac{x^3+x+1}{x^2-1} = x + \frac{3}{2(x-1)} + \frac{1}{2(x+1)},$ $\begin{aligned} \int f(x)\,dx &= \int x\,dx + \frac32\int\frac{dx}{x-1} \\ &\quad + \frac12\int\frac{dx}{x+1} \\ &= \frac{x^2}{2} + \frac32\log|x-1| \\ &\quad + \frac12\log|x+1| + C. \end{aligned}$
Integral of a sum, constants outside (Properties (III), (IV), Section 7.2.1). The polynomial part $x$ gives $\frac{x^2}{2}$ by the power rule; each simple fraction gives a logarithm.

Standard formulae (i) and (xii), Section 7.2. Basics: logarithms and 1/x
Check: differentiate the answer
$\begin{aligned} F'(x) &= x + \frac{3}{2(x-1)} + \frac{1}{2(x+1)} \\ &= x + \frac{3(x+1) + (x-1)}{2(x^2-1)} \\ &= x + \frac{4x+2}{2(x^2-1)} = x + \frac{2x+1}{x^2-1} \\ &= \frac{x^3 - x + 2x + 1}{x^2-1} = \frac{x^3+x+1}{x^2-1}. \ \checkmark \end{aligned}$
The derivative returns the original improper fraction. The answer is valid on each interval avoiding $x = \pm 1$ ⁠.

Power rule and $\frac{d}{dx}\log|x| = \frac1x$ ⁠, Section 7.2. Basics: integration undoes differentiation

Result

For $x \ne \pm 1$ ⁠:

\begin{aligned} &\int\frac{x^3+x+1}{x^2-1}\,dx \\ &= \frac{x^2}{2} + \frac12\log|x+1| + \frac32\log|x-1| + C. \end{aligned}

(Textbook answer: $\frac{x^2}{2} + \frac12\log|x+1| + \frac32\log|x-1| + C$ ⁠.)

Check it with numbers

Decomposition at $x = 2$ ⁠: the integrand is $\frac{8 + 2 + 1}{3} = \frac{11}{3}$ ⁠, and $2 + \frac32 + \frac{1/2}{3} = 2 + \frac32 + \frac16 = \frac{11}{3}$ ⁠.

Area check on $[2, 3]$ ⁠, where the integrand is continuous: with $F$ our answer,

\begin{aligned} F(3) - F(2) &= \tfrac52 + \tfrac52\log 2 - \tfrac12\log 3 \\ &\approx 3.6835618. \end{aligned}

Numerical integration from 2 to 3 also gives $3.6835618$ ⁠.

Common mistakes

Splitting without dividing. $\frac{A}{x-1} + \frac{B}{x+1}$ alone cannot equal an improper fraction; the $x$ part would be lost.
Wrong remainder. $x(x^2-1) = x^3 - x$ ⁠, so the remainder is $x + 1 - (-x) = 2x + 1$ ⁠, not $1$ ⁠.
Forgetting to integrate the quotient $x$ ⁠, which contributes $\frac{x^2}{2}$ ⁠.

What to take away

Degree of numerator $\ge$ degree of denominator: divide first.
Integrate the quotient as a polynomial and the remainder fraction by partial fractions.

Every fact used, and where it comes from

Fact	Used in	Source
Improper fraction = polynomial + proper fraction	Steps 1, 2	Section 7.5; Example 12
$x^2 - 1 = (x-1)(x+1)$	Step 3	Algebra (Basics)
Form $\frac{A}{x-a} + \frac{B}{x-b}$	Step 3	Table 7.2, row 1
The decomposition is an identity	Steps 3, 4	Remark after Example 11, Section 7.5
Equal polynomials have equal coefficients	Step 4	Algebra (Basics)
Integral of a sum; constants outside	Step 5	Properties (III), (IV), Section 7.2.1
$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ ⁠; $\int\frac{dt}{t} = \log\|t\| + C$	Step 5	Standard formulae (i), (xii), Section 7.2
Derivatives of powers and of $\log\|x\|$	Step 6	Section 7.2

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Exercise 7.5 · Q12: ∫x3+x+1x2−1 dx\int\frac{x^3+x+1}{x^2-1}\,dx∫x2−1x3+x+1​dx

Basics you need

Solution

The fraction is improper

Divide x3+x+1x^3 + x + 1x3+x+1 by x2−1x^2 - 1x2−1

Split the proper fraction

Find AAA and BBB

Integrate

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.5 · Q12: $\int\frac{x^3+x+1}{x^2-1}\,dx$

Divide $x^3 + x + 1$ by $x^2 - 1$

Find $A$ and $B$