Chapter 7 · Integrals

Exercise 7.6 · Q5: $\int x\log 2x\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

5. $x\log 2x$

Exam answer

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

Taking $\log 2x$ as the first function (its derivative is $\frac{2}{2x} = \frac1x$ ⁠), by parts:

\begin{aligned} I &= \log 2x\cdot\frac{x^2}{2} - \int\frac1x\cdot\frac{x^2}{2}\,dx \\ &= \frac{x^2}{2}\log 2x - \frac12\int x\,dx \\ &= \frac{x^2}{2}\log 2x - \frac12\cdot\frac{x^2}{2} + C \end{aligned}

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

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

The one idea this problem is built on

Exactly as in Question 4, the logarithm is the function to differentiate. The only new point: the derivative of $\log 2x$ is still $\frac1x$ ⁠, because the chain rule's factor 2 cancels the 2 inside. So the working is the same as Question 4 with $\log 2x$ in place of $\log x$ ⁠.

Given, to find, and the plan

Given: The function $x\log 2x$ ⁠. It is defined only for $2x > 0$ ⁠, that is $x > 0$ ⁠.
To find: $\int x\log 2x\,dx$ on $x > 0$ ⁠.
Plan: (1) Name $u$ and $\frac{dv}{dx}$ ⁠. (2) Find $\frac{du}{dx}$ (chain rule) and $v$ ⁠. (3) Apply the formula. (4) Finish. (5) Cross-check with Question 4. (6) Differentiate back.

Basics you need

1Integration by parts

Section 7.6, equation (1):

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx,

with $u$ the first function (differentiated) and $\frac{dv}{dx}$ the second (integrated to give $v$ ⁠). Lesson: integration by parts

2Which factor should be $u$ ⁠?

Remark (iii) after Example 17: when the other function is logarithmic, the logarithm is the first function. Memory aid ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential; not in the book): L comes before A.

3Chain rule for $\log 2x$

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

4Log of a product

$\log 2x = \log 2 + \log x$ for $x > 0$ ⁠, used in the cross-check. Basics: logarithms and 1/x

Solution

Name $u$ and $\frac{dv}{dx}$
$\begin{aligned} u &= \log 2x \quad (\text{first}), \\ \frac{dv}{dx} &= x \quad (\text{second}). \end{aligned}$
ILATE: Logarithmic before Algebraic. Differentiating $\log 2x$ gives a simple algebraic $\frac1x$ ⁠; integrating $x$ is the easy power rule. The other way round we would first need $\int\log 2x\,dx$ ⁠, which is itself a by-parts problem.

Remark (iii) after Example 17, Section 7.6. Lesson: integration by parts
Find $\frac{du}{dx}$ and $v$
$\frac{du}{dx} = \frac{1}{2x}\cdot 2 = \frac1x, \qquad v = \int x\,dx = \frac{x^2}{2}.$
Chain rule: the outer function is $\log$ ⁠, with derivative $\frac{1}{2x}$ at $2x$ ⁠; the inner function is $2x$ ⁠, with derivative 2. Their product is $\frac1x$ ⁠. No constant in $v$ (Remark (ii) after Example 17).

Chain rule (Chapter 5); standard formula (i). Basics: logarithms and 1/x
Apply the formula
$\begin{aligned} \int x\log 2x\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= \frac{x^2}{2}\log 2x - \int\frac{x^2}{2}\cdot\frac1x\,dx \\ &= \frac{x^2}{2}\log 2x - \frac12\int x\,dx. \end{aligned}$
$\frac{x^2}{2}\cdot\frac1x = \frac x2$ (as $x \ne 0$ ⁠): the logarithm is gone from the remaining integral.

Section 7.6, equation (1). Lesson: integration by parts
Finish the remaining integral
$\int x\log 2x\,dx = \frac{x^2}{2}\log 2x - \frac{x^2}{4} + C.$
$\frac12\int x\,dx = \frac12\cdot\frac{x^2}{2} = \frac{x^2}{4}$ ⁠, by the power rule. One constant $C$ at the end.

Standard formula (i), Section 7.2. Basics: powers of x
Cross-check with Question 4

For $x > 0$ ⁠, $\log 2x = \log 2 + \log x$ ⁠, so
$\begin{aligned} \int x\log 2x\,dx &= \log 2\int x\,dx + \int x\log x\,dx \\ &= \frac{x^2}{2}\log 2 + \frac{x^2}{2}\log x - \frac{x^2}{4} + C \\ &= \frac{x^2}{2}\log 2x - \frac{x^2}{4} + C. \end{aligned}$
The second route uses Question 4's answer and the law $\log a + \log b = \log ab$ ⁠. Both routes give the same function, which is a strong check.

Laws of logarithms; Question 4. Exercise 7.6 Q4
Check: differentiate the answer
$\begin{aligned} &\frac{d}{dx}\Big(\frac{x^2}{2}\log 2x - \frac{x^2}{4}\Big) \\ &= x\log 2x + \frac{x^2}{2}\cdot\frac1x - \frac x2 \\ &= x\log 2x. \end{aligned}$
Product rule on the first term (with $\frac{d}{dx}\log 2x = \frac1x$ ⁠); the $\frac x2$ terms cancel and the integrand is left, for every $x > 0$ ⁠.

Product rule and chain rule. Basics: integration undoes differentiation

Result

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

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

Check it with numbers

With $F(x) = \frac{x^2}{2}\log 2x - \frac{x^2}{4}$ ⁠, on $\left[\frac12, \frac e2\right]$ ⁠, where $\log 1 = 0$ and $\log e = 1$ ⁠:

\begin{aligned} F\left(\tfrac e2\right) - F\left(\tfrac12\right) &= \left(\frac{e^2}{8} - \frac{e^2}{16}\right) - \left(0 - \frac{1}{16}\right) \\ &= \frac{e^2 + 1}{16} = 0.5243160062\ldots \end{aligned}

Numerical integration of $x\log 2x$ over $\left[\frac12, \frac e2\right]$ gives $0.5243160062\ldots$ ⁠, the same.

Common mistakes

$\frac{d}{dx}\log 2x = \frac{1}{2x}$ ⁠. This forgets the inner derivative 2; the correct derivative is $\frac1x$ ⁠.
Changing $\log 2x$ to $\log x$ in the answer. They differ by the constant $\log 2$ ⁠, but $\frac{x^2}{2}\log 2$ is not a constant, so the 2 must stay (or be handled as in Step 5).
Forgetting the domain $x > 0$ ⁠.

What to take away

$\frac{d}{dx}\log kx = \frac1x$ for any constant $k > 0$ (and $x > 0$ ⁠).
The method of Question 4 works unchanged: logarithm as $u$ ⁠, power of $x$ as $\frac{dv}{dx}$ ⁠.
A second route (here, splitting the logarithm) is a cheap and strong check.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts	Step 3	Section 7.6, equation (1) (Lesson: integration by parts)
Logarithm as the first function (ILATE)	Step 1	Remark (iii) after Example 17, Section 7.6
No constant needed in $v$	Step 2	Remark (ii) after Example 17, Section 7.6
$\frac{d}{dx}\log x = \frac1x$ ⁠; chain rule	Steps 2, 6	Standard formula (xii), Section 7.2; Chapter 5, Section 5.3.1
Power rule $\int x\,dx = \frac{x^2}{2} + C$	Steps 2, 4, 5	Standard formula (i), Section 7.2
$\log ab = \log a + \log b$	Step 5	Laws of logarithms (Chapter 5, Section 5.4)
Product rule	Step 6	Chapter 5, Section 5.3

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Exercise 7.6 · Q5: ∫xlog⁡2x dx\int x\log 2x\,dx∫xlog2xdx

Basics you need

Solution

Name uuu and dvdx\frac{dv}{dx}dxdv​

Find dudx\frac{du}{dx}dxdu​ and vvv

Apply the formula

Finish the remaining integral

Cross-check with Question 4

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q5: $\int x\log 2x\,dx$

Name $u$ and $\frac{dv}{dx}$

Find $\frac{du}{dx}$ and $v$