Chapter 7 · Integrals

Exercise 7.6 · Q19: $\int e^x\left(\frac1x-\frac1{x^2}\right)dx$

Exercise 7.6Integral of the type eˣ[f + f′] · Section 7.6.1

The question

Integrate the functions in Exercises 1 to 22.

19.

e^x\left(\frac1x - \frac{1}{x^2}\right)

Exam answer

Let $I = \displaystyle\int e^x\left(\frac1x - \frac{1}{x^2}\right)dx$ ⁠.

Let $f(x) = \frac1x$ ⁠. Then $f'(x) = -\frac{1}{x^2}$ ⁠.

I = \int e^x\left[f(x) + f'(x)\right]dx

Using $\displaystyle\int e^x\left[f(x) + f'(x)\right]dx$ $= e^xf(x) + C$ ⁠,

$\therefore\ I = \dfrac{e^x}{x} + C$ ⁠.

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

The one idea this problem is built on

The derivative of $\frac1x$ is $-\frac{1}{x^2}$ ⁠. So the bracket is already "a function plus its derivative", $f + f'$ with $f(x) = \frac1x$ ⁠, and Section 7.6.1 gives $\int e^x[f + f']\,dx = e^x f + C$ ⁠.

Given, to find, and the plan

Given: $e^x\left(\frac1x - \frac1{x^2}\right)$ ⁠, defined for $x \ne 0$ ⁠.
To find: $\int e^x\left(\frac1x - \frac1{x^2}\right)dx$ ⁠.
Plan: (1) Identify $f$ and check $f'$ ⁠. (2) Apply the rule. (3) Differentiate back.

Basics you need

1The rule $\int e^x[f + f']\,dx = e^x f + C$

Section 7.6.1; valid on any interval where $f$ is differentiable. Lesson: integration by parts

2Derivative of $\frac1x$

$\frac{d}{dx}x^{-1} = -x^{-2} = -\frac{1}{x^2}$ for $x \ne 0$ (power rule). Basics: powers of x

Solution

Identify $f$ and check $f'$

Take $f(x) = \frac1x = x^{-1}$ ⁠. By the power rule,
$f'(x) = -1\cdot x^{-2} = -\frac{1}{x^2}.$
So $\frac1x - \frac1{x^2} = f(x) + f'(x)$ ⁠.

Of the two terms, one must be the derivative of the other. Differentiating $\frac1x$ gives $-\frac1{x^2}$ ⁠, with the minus sign that appears in the question, so $f = \frac1x$ ⁠. (Going the other way, the derivative of $-\frac{1}{x^2}$ is $\frac{2}{x^3}$ ⁠, not $\frac1x$ ⁠.)

$\frac{d}{dx}x^n = nx^{n-1}$ with $n = -1$ ⁠. Basics: powers of x
Apply the rule
$\int e^x\left(\frac1x - \frac1{x^2}\right)dx = \frac{e^x}{x} + C.$
$\int e^x[f + f']\,dx = e^x f + C$ with $f = \frac1x$ ⁠. Since $\frac1x$ is differentiable on $(-\infty, 0)$ and on $(0, \infty)$ ⁠, the answer holds on each of these intervals (the constant may differ between them).

Section 7.6.1. Lesson: integration by parts
Check: differentiate the answer
$\begin{aligned} \frac{d}{dx}\left[\frac{e^x}{x}\right] &= \frac{e^x\cdot x - e^x\cdot 1}{x^2} \\ &= \frac{e^x}{x} - \frac{e^x}{x^2} = e^x\left(\frac1x - \frac1{x^2}\right). \end{aligned}$
Quotient rule with top $e^x$ (derivative $e^x$ ⁠) and bottom $x$ (derivative 1); then split the fraction. The result is the integrand for every $x \ne 0$ ⁠.

Quotient rule (Chapter 5, Section 5.3). Basics: integration undoes differentiation

Result

\int e^x\left(\frac1x - \frac1{x^2}\right)dx = \frac{e^x}{x} + C,

valid on any interval not containing $0$ ⁠. (Textbook answer: $\frac{e^x}{x} + C$ ⁠, the same.)

Check it with numbers

On $[1, 2]$ ⁠, which does not contain 0:

\begin{aligned} F(2) - F(1) &= \frac{e^2}{2} - e \\ &= 0.9762462210\ldots \end{aligned}

Numerical integration of $e^x\left(\frac1x - \frac1{x^2}\right)$ over $[1, 2]$ gives $0.9762462210\ldots$ ⁠, the same.

Common mistakes

Splitting into $\int\frac{e^x}{x}\,dx - \int\frac{e^x}{x^2}\,dx$ and trying each separately: neither has an answer in elementary functions on its own (a known result, not proved in the book; Remark (ii) after Example 4 notes that such integrals exist); only the combination does.
Choosing $f = -\frac1{x^2}$ ⁠. Its derivative is $\frac{2}{x^3}$ ⁠, not $\frac1x$ ⁠.
Forgetting $x \ne 0$ ⁠.

What to take away

$\int e^x\left(\frac1x - \frac1{x^2}\right)dx = \frac{e^x}{x} + C$ ⁠: a direct case of $\int e^x[f + f']\,dx = e^x f + C$ ⁠.
When two terms in the bracket differ by one power of $x$ with the right coefficient, test whether one is the derivative of the other.

Every fact used, and where it comes from

Fact	Used in	Source
$\int e^x[f + f']\,dx = e^x f + C$	Step 2	Section 7.6.1 (Lesson: integration by parts)
$\frac{d}{dx}x^n = nx^{n-1}$	Step 1	Standard formula (i), Section 7.2; Chapter 5
Quotient rule	Step 3	Chapter 5, Section 5.3

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Exercise 7.6 · Q19: ∫ex(1x−1x2)dx\int e^x\left(\frac1x-\frac1{x^2}\right)dx∫ex(x1​−x21​)dx

Basics you need

Solution

Identify fff and check f′f'f′

Apply the rule

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q19: $\int e^x\left(\frac1x-\frac1{x^2}\right)dx$

Identify $f$ and check $f'$