Chapter 7 · Integrals

Exercise 7.6 · Q14: $\int x(\log x)^2\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

14.

x\,(\log x)^2

Exam answer

Let $I = \displaystyle\int x(\log x)^2\,dx$ ⁠.

Taking $(\log x)^2$ as the first function, by parts:

\begin{aligned} I &= (\log x)^2\cdot\frac{x^2}{2} \\ &\quad - \int\frac{2\log x}{x}\cdot\frac{x^2}{2}\,dx \\ &= \frac{x^2}{2}(\log x)^2 - \int x\log x\,dx \end{aligned}

Again by parts, taking $\log x$ as the first function:

\begin{aligned} &\int x\log x\,dx \\ &= \log x\cdot\frac{x^2}{2} - \int\frac1x\cdot\frac{x^2}{2}\,dx \\ &= \frac{x^2}{2}\log x - \frac{x^2}{4} \end{aligned}

\begin{aligned} \therefore\ I &= \frac{x^2}{2}(\log x)^2 \\ &\quad - \left(\frac{x^2}{2}\log x - \frac{x^2}{4}\right) + C \end{aligned}

$\therefore\ I = \dfrac{x^2}{2}(\log x)^2 - \dfrac{x^2}{2}\log x$ $+ \dfrac{x^2}{4} + C$ ⁠.

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

The one idea this problem is built on

Integrating by parts with $(\log x)^2$ as the first function replaces it by its derivative $\frac{2\log x}{x}$ ⁠: the power of the logarithm drops from 2 to 1, and the $\frac1x$ cancels against the $x^2$ that comes from integrating $x$ ⁠. One more round of parts removes the last logarithm. Each round lowers the power of $\log x$ by one.

Given, to find, and the plan

Given: $x(\log x)^2$ ⁠, defined for $x > 0$ (where $\log x$ is defined).
To find: $\int x(\log x)^2\,dx$ ⁠.
Plan: (1) Choose $(\log x)^2$ first, $x$ second. (2) Apply parts; the new integral is $\int x\log x\,dx$ ⁠. (3) Integrate that by parts. (4) Combine. (5) Differentiate back.

Basics you need

1Integration by parts

$\int f g\,dx = f\int g\,dx - \int\big[f'\int g\,dx\big]dx$ (Section 7.6). Lesson: integration by parts

2Derivative of $\log x$ and the chain rule

$\frac{d}{dx}\log x = \frac1x$ for $x > 0$ ⁠, so $\frac{d}{dx}(\log x)^2 = 2\log x\cdot\frac1x$ ⁠. Basics: logarithms and 1/x

3Power rule

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for $n \ne -1$ (standard formula (i)). Basics: powers of x

Solution

Choose the first and second functions

Take
$f(x) = (\log x)^2\ (\text{first}),\qquad g(x) = x\ (\text{second}).$
Then, by the chain rule, $f'(x) = \frac{2\log x}{x}$ ⁠, and $\int x\,dx = \frac{x^2}{2}$ ⁠.

We cannot integrate $(\log x)^2$ directly, so it cannot be the second function. Differentiating it, on the other hand, lowers the power of the logarithm. This matches Remark (iii) in Section 7.6: a logarithmic function is taken as the first function even when the other factor is a power of $x$ ⁠. (A common memory aid, not in the book, is ILATE: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. L comes before A, so $(\log x)^2$ is $u$ and $x$ is $\frac{dv}{dx}$ ⁠.)

$\frac{d}{dx}(\log x)^2 = 2(\log x)\cdot\frac1x$ (chain rule, Section 5.3.1). Basics: logarithms and 1/x
First round of integration by parts
$\begin{aligned} \int x(\log x)^2\,dx &= (\log x)^2\cdot\frac{x^2}{2} - \int\frac{2\log x}{x}\cdot\frac{x^2}{2}\,dx \\ &= \frac{x^2}{2}(\log x)^2 - \int x\log x\,dx. \end{aligned}$
Inside the new integral, $\frac{2}{x}\cdot\frac{x^2}{2} = x$ ⁠, so the logarithm now appears only to the first power. That is progress: $\int x\log x\,dx$ is Exercise 7.6 Question 4.

Parts formula with $f = (\log x)^2$ ⁠, $g = x$ ⁠. Lesson: integration by parts
Second round: $\int x\log x\,dx$

Again take the logarithm as the first function: $\log x$ first, $x$ second. Then the derivative of the first is $\frac1x$ and the integral of the second is $\frac{x^2}{2}$ ⁠:
$\begin{aligned} \int x\log x\,dx &= \log x\cdot\frac{x^2}{2} - \int\frac1x\cdot\frac{x^2}{2}\,dx \\ &= \frac{x^2}{2}\log x - \int\frac{x}{2}\,dx \\ &= \frac{x^2}{2}\log x - \frac{x^2}{4}. \end{aligned}$
Same reasoning as in Step 1. After this round no logarithm is left inside the integral, and $\int\frac x2\,dx = \frac{x^2}{4}$ by the power rule.

$\int x\,dx = \frac{x^2}{2}$ (formula (i)). Basics: powers of x
Combine

Substitute Step 3 into Step 2, taking care with the minus sign in front of the bracket:
$\begin{aligned} \int x(\log x)^2\,dx &= \frac{x^2}{2}(\log x)^2 - \left[\frac{x^2}{2}\log x - \frac{x^2}{4}\right] \\ &= \frac{x^2}{2}(\log x)^2 - \frac{x^2}{2}\log x + \frac{x^2}{4} + C, \end{aligned}$
for $x > 0$ ⁠.

The minus in front of the bracket changes $-\frac{x^2}{4}$ into $+\frac{x^2}{4}$ ⁠. The answer is valid for $x > 0$ ⁠, the domain of $\log x$ ⁠.

Antiderivatives differ by a constant: one $C$ at the end (Section 7.2). Basics: integration undoes differentiation
Check: differentiate the answer

Differentiate the three terms separately (product rule on each):
$\begin{aligned} \frac{d}{dx}\left[\frac{x^2}{2}(\log x)^2\right] &= x(\log x)^2 + x\log x, \\ \frac{d}{dx}\left[-\frac{x^2}{2}\log x\right] &= -x\log x - \frac{x}{2}, \\ \frac{d}{dx}\left[\frac{x^2}{4}\right] &= \frac{x}{2}. \end{aligned}$
Adding, the $x\log x$ terms cancel and the $\frac x2$ terms cancel, leaving $x(\log x)^2$ ⁠, the integrand, for every $x > 0$ ⁠.

First line: $x\cdot(\log x)^2 + \frac{x^2}{2}\cdot\frac{2\log x}{x}$ ⁠. Second line: $-\big(x\log x + \frac{x^2}{2}\cdot\frac1x\big)$ ⁠.

Product rule; $\frac{d}{dx}\log x = \frac1x$ ⁠. Basics: integration undoes differentiation

Result

\begin{aligned} &\int x(\log x)^2\,dx \\ &= \frac{x^2}{2}(\log x)^2 - \frac{x^2}{2}\log x + \frac{x^2}{4} + C, \end{aligned}

valid for $x > 0$ ⁠. (Textbook answer: $\frac{x^2}{2}(\log x)^2 - \frac{x^2}{2}\log x + \frac{x^2}{4} + C$ ⁠, the same.)

Check it with numbers

On $[1, e]$ ⁠, with $F$ the answer without $C$ ⁠: $\log e = 1$ and $\log 1 = 0$ ⁠, so

\begin{aligned} F(e) - F(1) &= \left(\frac{e^2}{2} - \frac{e^2}{2} + \frac{e^2}{4}\right) - \frac14 \\ &= \frac{e^2 - 1}{4} = 1.5972640247\ldots \end{aligned}

Numerical integration of $x(\log x)^2$ over $[1, e]$ gives $1.5972640247\ldots$ ⁠, the same.

Common mistakes

Taking $x$ as the first function. Then the second function $(\log x)^2$ has to be integrated, which is harder than the original problem.
Forgetting the chain rule: $\frac{d}{dx}(\log x)^2$ is $\frac{2\log x}{x}$ ⁠, not $\frac{2}{x}$ or $2\log x$ ⁠.
Sign slip when combining. $-\big[\frac{x^2}{2}\log x - \frac{x^2}{4}\big] = -\frac{x^2}{2}\log x + \frac{x^2}{4}$ ⁠.

What to take away

For $x^m(\log x)^k$ ⁠, keep the logarithm as the first function; each round of parts lowers $k$ by one.
Reuse earlier results: the second round here is exactly Exercise 7.6 Question 4.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts	Steps 2, 3	Section 7.6 (Lesson: integration by parts)
Logarithmic function taken as the first function	Steps 1, 3	Remark (iii), Section 7.6
$\frac{d}{dx}\log x = \frac1x$ for $x > 0$ ⁠; chain rule	Steps 1, 3, 5	Chapter 5, Sections 5.4 and 5.3.1
$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$	Steps 1, 3	Standard formula (i), Section 7.2
Product rule	Step 5	Chapter 5, Section 5.3

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Exercise 7.6 · Q14: ∫x(log⁡x)2 dx\int x(\log x)^2\,dx∫x(logx)2dx

Basics you need

Solution

Choose the first and second functions

First round of integration by parts

Second round: ∫xlog⁡x dx\int x\log x\,dx∫xlogxdx

Combine

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q14: $\int x(\log x)^2\,dx$

Second round: $\int x\log x\,dx$