Chapter 7 · Integrals

Exercise 7.1 · Q11: $\int \frac{x^3 + 5x^2 - 4}{x^2}\,dx$

Exercise 7.1Properties of indefinite integrals · Section 7.2.1

The question

Find the following integral:

\int\frac{x^3 + 5x^2 - 4}{x^2}\,dx

Exam answer

Let $I = \displaystyle\int\frac{x^3 + 5x^2 - 4}{x^2}\,dx$ ⁠.

\begin{aligned} I &= \int\left(x + 5 - 4x^{-2}\right)dx \\ &= \frac{x^2}{2} + 5x - 4\cdot\frac{x^{-1}}{-1} + C \end{aligned}

$\therefore\ I = \dfrac{x^2}{2} + 5x + \dfrac4x + C$ ⁠.

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

The one idea this problem is built on

The denominator is a single power of $x$ ⁠, so the fraction splits term by term: divide each term of the numerator by $x^2$ ⁠. That gives $x + 5 - 4x^{-2}$ ⁠, three powers of $x$ ⁠, each handled by the power rule. The negative power $x^{-2}$ is allowed (only $x^{-1}$ is excluded).

Given, to find, and the plan

Given: The integrand $\frac{x^3 + 5x^2 - 4}{x^2}$ ⁠, defined for every real $x \ne 0$ ⁠.
To find: The indefinite integral.
Plan: (1) Split the fraction over the single-term denominator. (2) Write each piece as a power of $x$ ⁠. (3) Power rule on each, including $x^{-2}$ ⁠. (4) Differentiate back to check. (5) State the domain.

Basics you need

1Splitting a fraction with a one-term denominator

$\frac{A + B + C}{D} = \frac{A}{D} + \frac{B}{D} + \frac{C}{D}$ whenever $D \ne 0$ ⁠. Basics: splitting fractions

2Laws of indices

$\frac{x^m}{x^n} = x^{m-n}$ and $\frac{1}{x^n} = x^{-n}$ for $x \ne 0$ ⁠. Basics: powers of x and the power rule

3Power rule, negative powers included

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for every real $n \ne -1$ (item (i), Section 7.2), so $n = -2$ is fine. Basics: the power rule

Solution

Split the fraction

For $x \ne 0$ ⁠, divide each term of the numerator by $x^2$ ⁠:
$\begin{aligned} \frac{x^3 + 5x^2 - 4}{x^2} &= \frac{x^3}{x^2} + \frac{5x^2}{x^2} - \frac{4}{x^2} \\ &= x + 5 - 4x^{-2}. \end{aligned}$
There is no quotient rule for integrals, but a fraction whose denominator is a single term splits into separate fractions. Each simplifies by the laws of indices: $\frac{x^3}{x^2} = x$ ⁠, $\frac{5x^2}{x^2} = 5$ ⁠, $\frac{4}{x^2} = 4x^{-2}$ ⁠. The book does the same with $\frac{x^3 - 1}{x^2}$ in Example 2 (i).

Splitting over a single-term denominator; $\frac{x^m}{x^n} = x^{m-n}$ ⁠. Basics: splitting fractions · Example 2
Integrate term by term
$\begin{aligned} \int x\,dx &= \frac{x^2}{2}, \\ \int 5\,dx &= 5x, \\ \int 4x^{-2}\,dx &= 4\cdot\frac{x^{-2+1}}{-2+1} = -4x^{-1}, \end{aligned}$
each up to a constant.

For $x^{-2}$ the power rule raises the exponent by one, to $-1$ ⁠, and divides by the new exponent $-1$ ⁠. This is legitimate because $n = -2 \ne -1$ ⁠. Dividing by $-1$ flips the sign, so $\int x^{-2}\,dx = -x^{-1}$ ⁠.

Standard formulae, item (i), Section 7.2; Properties (III) to (V), Section 7.2.1. Basics: the power rule
Assemble the answer

The integrand is $x + 5 - 4x^{-2}$ ⁠, so we subtract the third piece:
$\begin{aligned} &\int\frac{x^3 + 5x^2 - 4}{x^2}\,dx \\ &= \frac{x^2}{2} + 5x - \left(-4x^{-1}\right) + C \\ &= \frac{x^2}{2} + 5x + \frac{4}{x} + C. \end{aligned}$
Two minus signs meet here: the $-$ in front of $4x^{-2}$ and the $-$ produced by dividing by $-1$ ⁠. Together they give $+\frac4x$ ⁠. This is the step where sign slips happen, so it is written out in full.

Note after Example 2: one constant of integration. Example 2
Check: differentiate the answer back
$\begin{aligned} &\frac{d}{dx}\left(\frac{x^2}{2} + 5x + 4x^{-1} + C\right) \\ &= x + 5 + 4(-1)x^{-2} \\ &= x + 5 - \frac{4}{x^2} \\ &= \frac{x^3 + 5x^2 - 4}{x^2}. \;\checkmark \end{aligned}$
In the last line we put everything over the common denominator $x^2$ ⁠: $x = \frac{x^3}{x^2}$ and $5 = \frac{5x^2}{x^2}$ ⁠. We get back the integrand exactly, for every $x \ne 0$ ⁠.

Property (I), Section 7.2.1; power rule for derivatives, Table 5.3 (Chapter 5). Basics: integration undoes differentiation
Where the answer is valid

The integrand and the answer both need $x \ne 0$ ⁠. The formula gives the anti derivatives on any interval not containing $0$ ⁠, such as $x > 0$ or $x < 0$ ⁠.

The term $\frac4x$ blows up at $0$ ⁠, just like the integrand. An anti derivative lives on an interval where the integrand is defined, so we work on one side of $0$ at a time.

Note after the standard formulae, Section 7.2. Lesson: integration as anti differentiation

Result

\int\frac{x^3 + 5x^2 - 4}{x^2}\,dx = \frac{x^2}{2} + 5x + \frac{4}{x} + C,

on any interval not containing $0$ ⁠. (Textbook answer: $\frac{x^2}{2} + 5x + \frac4x + C$ ⁠.)

Check it with numbers

At $x = 2$ with $h = 0.001$ ⁠, for $F(x) = \frac{x^2}{2} + 5x + \frac4x$ ⁠:

\begin{aligned} &\frac{F(2.001) - F(1.999)}{0.002} \\ &= \frac{14.006000999750 - 13.994001000250}{0.002} \\ &\approx 5.99999975. \end{aligned}

The integrand at $x = 2$ is $\frac{8 + 20 - 4}{4} = \frac{24}{4} = 6$ ⁠. ✓

Common mistakes

Wrong sign on $\frac4x$ ⁠. $\int(-4x^{-2})\,dx = -4\cdot\frac{x^{-1}}{-1} = +\frac4x$ ⁠. Answers with $-\frac4x$ lose one of the two minus signs.
Integrating top and bottom separately. $\frac{\int(x^3 + 5x^2 - 4)\,dx}{\int x^2\,dx}$ has nothing to do with the integral of the fraction.
Splitting with a sum in the denominator. Splitting works here because the denominator $x^2$ is a single term. $\frac{1}{x^2 + 1}$ does not split as $\frac{1}{x^2} + 1$ ⁠.

What to take away

A fraction over a single power of $x$ splits into powers of $x$ ⁠; then use the power rule.
Negative powers are fine in the power rule; only $x^{-1}$ needs the logarithm.
Track signs carefully when dividing by a negative new exponent.

Every fact used, and where it comes from

Fact	Used in	Source
Split over a single-term denominator; $\frac{x^m}{x^n} = x^{m-n}$	Step 1	Algebra (laws of indices); same move as Example 2 (i)
$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ ⁠, $n \ne -1$	Step 2	Standard formulae, item (i), Section 7.2
Integrate term by term	Steps 2, 3	Properties (III) to (V), Section 7.2.1
One constant of integration	Step 3	Note after Example 2, Section 7.2.1
$\frac{d}{dx}x^n = nx^{n-1}$	Step 4	Table 5.3, Section 5.3 (Chapter 5)
Keep the interval in mind	Step 5	Note after the standard formulae, Section 7.2

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Exercise 7.1 · Q11: ∫x3+5x2−4x2 dx\int \frac{x^3 + 5x^2 - 4}{x^2}\,dx∫x2x3+5x2−4​dx

Basics you need

Solution

Split the fraction

Integrate term by term

Assemble the answer

Check: differentiate the answer back

Where the answer is valid

Every fact used, and where it comes from

Exercise 7.1 · Q11: $\int \frac{x^3 + 5x^2 - 4}{x^2}\,dx$