Chapter 7 · Integrals

Exercise 7.2 · Q3: $\int \frac{dx}{x+x\log x}$

Exercise 7.2Integration by substitution · Section 7.3.1

The question

Integrate the function:

\frac{1}{x + x\log x}

Exam answer

Let $I = \displaystyle\int\frac{dx}{x + x\log x}$ $= \displaystyle\int\frac{dx}{x(1 + \log x)}$ ⁠.

Put $1 + \log x = t$ ⁠. Then $\frac1x\,dx = dt$ ⁠.

\therefore\ I = \int\frac{dt}{t} = \log|t| + C

$\therefore\ I = \log\left|1 + \log x\right| + C$ ⁠.

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

The one idea this problem is built on

Take out the common factor $x$ ⁠: the denominator is $x(1+\log x)$ ⁠. The integrand is then $\frac{1}{1+\log x}$ times $\frac1x$ ⁠, and $\frac1x$ is the derivative of $1+\log x$ ⁠. So $t = 1+\log x$ turns it into $\int\frac{dt}{t} = \log|t|$ ⁠.

Given, to find, and the plan

Given: $f(x) = \dfrac{1}{x + x\log x}$ ⁠. It needs $x > 0$ (for $\log x$ ⁠) and a non-zero denominator. Since $x + x\log x = x(1+\log x)$ and $x \ne 0$ ⁠, the denominator is zero only when $\log x = -1$ ⁠, that is $x = e^{-1} = \frac1e$ ⁠. So $f$ is defined for $x > 0$ ⁠, $x \ne \frac1e$ ⁠.
To find: $\displaystyle\int \frac{dx}{x + x\log x}$ ⁠.
Plan: (1) Factor the denominator. (2) Substitute $t = 1+\log x$ ⁠. (3) Integrate $\frac1t$ ⁠. (4) Return to $x$ ⁠. (5) Differentiate to check.

Basics you need

1Taking out a common factor

$x + x\log x = x\cdot 1 + x\cdot\log x$ ⁠, and taking out the common factor $x$ gives $x(1+\log x)$ ⁠.

2Substitution

Put $t = g(x)$ ⁠, so $dt = g'(x)\,dx$ (Section 7.3.1). Basics: substitution

3Logs

$\frac{d}{dx}\log x = \frac1x$ for $x>0$ ⁠, and $\int\frac{dt}{t} = \log|t| + C$ for $t \ne 0$ (standard formula (xii), Section 7.2). Basics: logarithms and 1/x

Solution

Factor the denominator
$\frac{1}{x + x\log x} = \frac{1}{x(1+\log x)} = \frac{1}{1+\log x}\cdot\frac{1}{x}.$
In its original form the integrand shows no pattern. Factoring reveals two pieces: a function of $1+\log x$ ⁠, and $\frac1x$ ⁠, which is the derivative of $1+\log x$ ⁠. That pairing is what substitution needs.
Substitute $t = 1+\log x$
$\begin{aligned} t &= 1 + \log x, \\ \frac{dt}{dx} &= 0 + \frac{1}{x}, \\ dt &= \frac{dx}{x}. \end{aligned}$
So
$\int \frac{1}{1+\log x}\cdot\frac{dx}{x} = \int \frac{dt}{t}.$
The first factor becomes $\frac1t$ and $\frac{dx}{x}$ is exactly $dt$ ⁠. On our domain $x \ne \frac1e$ ⁠, so $t = 1+\log x \ne 0$ ⁠, and $\frac1t$ is defined.

Substitution, Section 7.3.1; $\frac{d}{dx}\log x = \frac1x$ (Chapter 5, Theorem 5). Lesson: integration by substitution
Integrate in $t$
$\int\frac{dt}{t} = \log|t| + C.$
Standard formula (xii) with the variable called $t$ ⁠. The absolute value matters here: $t = 1+\log x$ is negative when $0 < x < \frac1e$ ⁠, and the log of a negative number is not defined.

Standard formula (xii), Section 7.2. Basics: logarithms and 1/x
Return to $x$
$\log|t| + C = \log\left|1+\log x\right| + C.$
Replace $t$ by $1+\log x$ ⁠. Keep the absolute value: it makes the answer valid on both pieces of the domain, $0 < x < \frac1e$ (where $1+\log x < 0$ ⁠) and $x > \frac1e$ (where $1+\log x > 0$ ⁠).
Check: differentiate the answer

Chain rule with inner $u = 1+\log x$ and outer $\log|u|$ ⁠, whose derivative is $\frac1u$ ⁠:
$\begin{aligned} \frac{d}{dx}\log\left|1+\log x\right| &= \frac{1}{1+\log x}\cdot\frac{1}{x} \\ &= \frac{1}{x(1+\log x)} \\ &= \frac{1}{x + x\log x}. \end{aligned}$
This is the integrand, on every part of the domain.

The derivative of $\log|u|$ is $\frac1u$ whether $u$ is positive or negative (Section 7.2 shows this for $\log|x|$ ⁠). So the check covers both intervals.

$\frac{d}{dx}\log|x| = \frac1x$ ⁠, $x \ne 0$ ⁠, Section 7.2; chain rule, Chapter 5, Theorem 4. Basics: integration undoes differentiation

Result

For $x > 0$ ⁠, $x \ne \frac1e$ ⁠,

\int \frac{dx}{x + x\log x} = \log\left|1+\log x\right| + C.

(Textbook answer: $\log|1+\log x| + C$ ⁠.) Strictly, the constant may be chosen separately on $(0, \frac1e)$ and on $(\frac1e, \infty)$ ⁠, because the integrand is not defined at $\frac1e$ ⁠.

Check it with numbers

Let $F(x) = \log|1+\log x|$ ⁠. At $x = 2$ the integrand is $\frac{1}{2+2\log 2} \approx 0.295308$ ⁠. With $h = 0.0001$ ⁠:

\begin{aligned} F(2.0001) &\approx 0.5266185638, \\ F(1.9999) &\approx 0.5265595022, \\ \frac{F(2.0001)-F(1.9999)}{0.0002} &\approx 0.295308. \end{aligned}

The slope of our answer equals the integrand.

Common mistakes

Splitting the fraction. $\frac{1}{x + x\log x}$ is not $\frac1x + \frac{1}{x\log x}$ ⁠. A sum in the denominator cannot be split.
Dropping the absolute value. $1+\log x$ is negative for $0 < x < \frac1e$ ⁠, so $\log(1+\log x)$ would be undefined there.
Choosing $t = \log x$ ⁠. It also works (giving $\int\frac{dt}{1+t}$ ⁠), but then a second step is needed; $t = 1+\log x$ does it at once.

What to take away

Factor first: a hidden common factor often reveals the derivative you need.
This is again the pattern $\int\frac{g'(x)}{g(x)}\,dx = \log|g(x)| + C$ ⁠, with $g(x) = 1+\log x$ ⁠.
State where the answer holds: here $x > 0$ and $x \ne \frac1e$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\log x = -1 \iff x = e^{-1}$	Given	Definition of the natural log (Chapter 5, Section 5.4)
$\frac{d}{dx}\log x = \frac1x$	Step 2	Chapter 5, Theorem 5
Substitution	Step 2	Section 7.3.1 (Lesson: substitution)
$\int\frac{dt}{t} = \log\|t\| + C$ ⁠; $\frac{d}{dx}\log\|x\| = \frac1x$	Steps 3, 5	Standard formula (xii), Section 7.2
Chain rule	Step 5	Chapter 5, Theorem 4

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Exercise 7.2 · Q3: ∫dxx+xlog⁡x\int \frac{dx}{x+x\log x}∫x+xlogxdx​

Basics you need

Solution

Factor the denominator

Substitute t=1+log⁡xt = 1+\log xt=1+logx

Integrate in ttt

Return to xxx

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.2 · Q3: $\int \frac{dx}{x+x\log x}$

Substitute $t = 1+\log x$

Integrate in $t$

Return to $x$