Chapter 7 · Integrals

Exercise 7.6 · Q4: $\int x\log x\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

4. $x\log x$

Exam answer

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

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

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

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

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

The one idea this problem is built on

Here the usual "power of $x$ first" rule is reversed. We have no formula yet for $\int\log x\,dx$ in a product, but the derivative of $\log x$ is the simple algebraic function $\frac1x$ ⁠. So $\log x$ is the function to differentiate, and $x$ is the one to integrate. After one use of the formula, $\frac1x$ cancels a power of $x$ and the remaining integral is a plain power.

Given, to find, and the plan

Given: The function $x\log x$ (natural logarithm). It is defined only for $x > 0$ ⁠, because $\log x$ is.
To find: $\int x\log x\,dx$ on $x > 0$ ⁠.
Plan: (1) Name $u$ and $\frac{dv}{dx}$ and explain why $\log x$ is $u$ ⁠. (2) Find $\frac{du}{dx}$ and $v$ ⁠. (3) Apply the formula and simplify. (4) Finish the remaining integral. (5) 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: a power of $x$ is usually first, but if the other function is logarithmic or inverse trigonometric, that one is first. Memory aid ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential; not in the book): L comes before A.

3Derivative of $\log x$

$\frac{d}{dx}\log x = \frac1x$ for $x > 0$ (standard formula (xii) read as a derivative). Basics: logarithms and 1/x

4Power rule

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

Solution

Name $u$ and $\frac{dv}{dx}$
$\begin{aligned} u &= \log x \quad (\text{first}), \\ \frac{dv}{dx} &= x \quad (\text{second}). \end{aligned}$
ILATE: $\log x$ is Logarithmic, $x$ is Algebraic, and L comes before A. The reason: if we tried $u = x$ ⁠, we would need $v = \int\log x\,dx$ first, which itself needs integration by parts (Example 18), and the new integral would be no simpler. With $u = \log x$ ⁠, differentiating gives the simple $\frac1x$ ⁠.

Remark (iii) after Example 17, Section 7.6. Lesson: integration by parts
Find $\frac{du}{dx}$ and $v$
$\frac{du}{dx} = \frac1x, \qquad v = \int x\,dx = \frac{x^2}{2}.$
The derivative of $\log x$ is $\frac1x$ (valid since $x > 0$ ⁠); $v$ comes from the power rule with $n = 1$ ⁠. No constant is added to $v$ (Remark (ii) after Example 17).

Standard formulae (xii) and (i), Section 7.2. Basics: logarithms and 1/x
Apply the formula
$\begin{aligned} \int x\log x\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= \frac{x^2}{2}\log x - \int\frac{x^2}{2}\cdot\frac1x\,dx \\ &= \frac{x^2}{2}\log x - \frac12\int x\,dx. \end{aligned}$
$\frac{x^2}{2}\cdot\frac1x = \frac x2$ because $x \ne 0$ ⁠: the $\frac1x$ from differentiating the logarithm cancels one power of $x$ ⁠. The logarithm has disappeared from the integral that is left.

Section 7.6, equation (1); constant multiple rule (Section 7.2.1). Lesson: integration by parts
Finish the remaining integral
$\begin{aligned} \int x\log x\,dx &= \frac{x^2}{2}\log x - \frac12\cdot\frac{x^2}{2} + C \\ &= \frac{x^2}{2}\log x - \frac{x^2}{4} + C. \end{aligned}$
$\int x\,dx = \frac{x^2}{2}$ by the power rule, multiplied by $\frac12$ ⁠. One constant $C$ at the end.

Standard formula (i), Section 7.2. Basics: powers of x
Check: differentiate the answer

Product rule on the first term:
$\begin{aligned} &\frac{d}{dx}\Big(\frac{x^2}{2}\log x - \frac{x^2}{4}\Big) \\ &= x\log x + \frac{x^2}{2}\cdot\frac1x - \frac{2x}{4} \\ &= x\log x + \frac x2 - \frac x2 = x\log x. \end{aligned}$
The derivative is exactly the integrand at every $x > 0$ ⁠.

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

Result

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

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

Check it with numbers

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

\begin{aligned} F(e) - F(1) &= \left(\frac{e^2}{2} - \frac{e^2}{4}\right) - \left(0 - \frac14\right) \\ &= \frac{e^2 + 1}{4} = 2.0972640247\ldots \end{aligned}

Numerical integration of $x\log x$ over $[1, e]$ gives $2.0972640247\ldots$ ⁠, the same.

Common mistakes

Taking $u = x$ ⁠. Then $v = \int\log x\,dx$ is needed first (Example 18), and the new integral contains the original one again: a longer, roundabout route.
Forgetting the domain. $\log x$ needs $x > 0$ ⁠; the answer is valid only there.
Last coefficient. $\frac12\int x\,dx = \frac{x^2}{4}$ ⁠, not $\frac{x^2}{2}$ ⁠.

What to take away

A logarithm times a power of $x$ ⁠: the logarithm is $u$ ⁠, because its derivative $\frac1x$ is algebraic and cancels against the power.
The same steps give, for $n \ne -1$ and $x > 0$ ⁠, $\int x^n\log x\,dx = \frac{x^{n+1}}{n+1}\log x - \frac{x^{n+1}}{(n+1)^2} + C.$ This question is $n = 1$ ⁠; Question 6 is $n = 2$ ⁠.
State the domain whenever a logarithm appears.

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; Example 18
No constant needed in $v$	Step 2	Remark (ii) after Example 17, Section 7.6
$\frac{d}{dx}\log x = \frac1x$ for $x > 0$	Steps 2, 5	Standard formula (xii), Section 7.2; Chapter 5
Power rule $\int x^n dx$	Steps 2, 4	Standard formula (i), Section 7.2
Constant multiple rule	Step 3	Property (IV), Section 7.2.1
Product rule	Step 5	Chapter 5, Section 5.3

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Exercise 7.6 · Q4: ∫xlog⁡x dx\int x\log x\,dx∫xlogxdx

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

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q4: $\int x\log x\,dx$

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

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