Chapter 7 · Integrals

Exercise 7.6 · Q6: $\int x^2\log x\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

6. $x^2\log x$

Exam answer

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

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

\begin{aligned} I &= \log x\cdot\frac{x^3}{3} - \int\frac1x\cdot\frac{x^3}{3}\,dx \\ &= \frac{x^3}{3}\log x - \frac13\int x^2\,dx \\ &= \frac{x^3}{3}\log x - \frac13\cdot\frac{x^3}{3} + C \end{aligned}

$\therefore\ I = \dfrac{x^3}{3}\log x - \dfrac{x^3}{9} + C$ ⁠.

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

The one idea this problem is built on

Same pattern as Question 4: differentiate the logarithm (it becomes $\frac1x$ ⁠) and integrate the power (it becomes $\frac{x^3}{3}$ ⁠). Their product $\frac{x^3}{3}\cdot\frac1x = \frac{x^2}{3}$ is a plain power, so one use of integration by parts finishes the job.

Given, to find, and the plan

Given: The function $x^2\log x$ ⁠, defined for $x > 0$ (where $\log x$ is defined).
To find: $\int x^2\log x\,dx$ on $x > 0$ ⁠.
Plan: (1) Name $u$ and $\frac{dv}{dx}$ ⁠. (2) Find $\frac{du}{dx}$ and $v$ ⁠. (3) Apply the formula. (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: 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 before A.

3Derivative of $\log x$

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

4Power rule

$\int x^2\,dx = \frac{x^3}{3} + C$ (standard formula (i) with $n = 2$ ⁠). Basics: powers of x

Solution

Name $u$ and $\frac{dv}{dx}$
$\begin{aligned} u &= \log x \quad (\text{first}), \\ \frac{dv}{dx} &= x^2 \quad (\text{second}). \end{aligned}$
ILATE: Logarithmic before Algebraic. If $x^2$ were $u$ we would need $\int\log x\,dx$ as $v$ ⁠, and the remaining integral would still contain a logarithm. With $u = \log x$ ⁠, the logarithm turns into the algebraic $\frac1x$ and disappears from the next integral.

Remark (iii) after Example 17, Section 7.6. Lesson: integration by parts
Find $\frac{du}{dx}$ and $v$
$\frac{du}{dx} = \frac1x, \qquad v = \int x^2\,dx = \frac{x^3}{3}.$
Derivative of the logarithm (valid as $x > 0$ ⁠), and the power rule with $n = 2$ ⁠. No constant in $v$ (Remark (ii) after Example 17).

Standard formulae (xii) and (i), Section 7.2. Basics: powers of x
Apply the formula
$\begin{aligned} \int x^2\log x\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= \frac{x^3}{3}\log x - \int\frac{x^3}{3}\cdot\frac1x\,dx \\ &= \frac{x^3}{3}\log x - \frac13\int x^2\,dx. \end{aligned}$
$\frac{x^3}{3}\cdot\frac1x = \frac{x^2}{3}$ because $x \ne 0$ ⁠. The constant $\frac13$ comes out of the integral.

Section 7.6, equation (1); constant multiple rule (Section 7.2.1). Lesson: integration by parts
Finish the remaining integral
$\begin{aligned} \int x^2\log x\,dx &= \frac{x^3}{3}\log x - \frac13\cdot\frac{x^3}{3} + C \\ &= \frac{x^3}{3}\log x - \frac{x^3}{9} + C. \end{aligned}$
$\int x^2\,dx = \frac{x^3}{3}$ once more, and $\frac13\cdot\frac13 = \frac19$ ⁠.

Standard formula (i), Section 7.2. Basics: powers of x
Check: differentiate the answer
$\begin{aligned} &\frac{d}{dx}\Big(\frac{x^3}{3}\log x - \frac{x^3}{9}\Big) \\ &= x^2\log x + \frac{x^3}{3}\cdot\frac1x - \frac{3x^2}{9} \\ &= x^2\log x + \frac{x^2}{3} - \frac{x^2}{3} = x^2\log x. \end{aligned}$
Product rule on the first term; the two $\frac{x^2}{3}$ terms cancel and the integrand remains, for every $x > 0$ ⁠.

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

Result

\int x^2\log x\,dx = \frac{x^3}{3}\log x - \frac{x^3}{9} + C,

valid for $x > 0$ ⁠. (Textbook answer: $\frac{x^3}{3}\log x - \frac{x^3}{9} + C$ ⁠, the same.)

Check it with numbers

With $F(x) = \frac{x^3}{3}\log x - \frac{x^3}{9}$ ⁠, on $[1, e]$ ⁠:

\begin{aligned} F(e) - F(1) &= \left(\frac{e^3}{3} - \frac{e^3}{9}\right) - \left(0 - \frac19\right) \\ &= \frac{2e^3 + 1}{9} = 4.5745637607\ldots \end{aligned}

Numerical integration of $x^2\log x$ over $[1, e]$ gives $4.5745637607\ldots$ ⁠, the same.

Common mistakes

Taking $u = x^2$ ⁠. The logarithm then stays inside the new integral.
Last coefficient. $\frac13\int x^2\,dx = \frac{x^3}{9}$ ⁠, not $\frac{x^3}{3}$ ⁠.
Forgetting the domain $x > 0$ ⁠.

What to take away

Logarithm times a power of $x$ ⁠: the logarithm is $u$ ⁠.
For $n \ne -1$ ⁠: $\int x^n\log x\,dx$ has the same shape of answer, $\frac{x^{n+1}}{n+1}\log x$ minus $\frac{x^{n+1}}{(n+1)^2}$ ⁠, plus $C$ (Question 4 is $n = 1$ ⁠).
Check the coefficients by differentiating: the cancelling pair is a good sign.

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
$\frac{d}{dx}\log x = \frac1x$ for $x > 0$	Steps 2, 5	Standard formula (xii), Section 7.2; Chapter 5
Power rule	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 · Q6: ∫x2log⁡x dx\int x^2\log x\,dx∫x2logxdx

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 · Q6: $\int x^2\log x\,dx$

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

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