Chapter 7 · Integrals

Exercise 7.6 · Q15: $\int(x^2+1)\log x\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

15.

(x^2 + 1)\log x

Exam answer

Let $I = \displaystyle\int(x^2 + 1)\log x\,dx$ ⁠.

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

\begin{aligned} I &= \log x\left(\frac{x^3}{3} + x\right) \\ &\quad - \int\frac1x\left(\frac{x^3}{3} + x\right)dx \\ &= \left(\frac{x^3}{3} + x\right)\log x \\ &\quad - \int\left(\frac{x^2}{3} + 1\right)dx \end{aligned}

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

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

The one idea this problem is built on

A polynomial times $\log x$ ⁠: integrate by parts with $\log x$ as the first function. Its derivative $\frac1x$ divides the integrated polynomial $\frac{x^3}{3} + x$ term by term, leaving a plain polynomial that the power rule finishes.

Given, to find, and the plan

Given: $(x^2+1)\log x$ ⁠, defined for $x > 0$ ⁠.
To find: $\int(x^2+1)\log x\,dx$ ⁠.
Plan: (1) Choose $\log x$ first, $x^2 + 1$ second. (2) Apply parts. (3) Simplify and integrate the leftover polynomial. (4) Write the answer. (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 $\log x$

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

3Power rule

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ ⁠, $n \ne -1$ (standard formula (i)); integrals of sums split (Section 7.2.1). Basics: powers of x

Solution

Choose the first and second functions

Take
$f(x) = \log x\ (\text{first}),\qquad g(x) = x^2 + 1\ (\text{second}).$
Then $f'(x) = \frac1x$ and $\int(x^2+1)\,dx = \frac{x^3}{3} + x$ ⁠.

Remark (iii), Section 7.6: a polynomial is usually the first function, but when the other factor is logarithmic, the logarithm is taken first. The reason is practical: $\log x$ has no simple integral, but its derivative $\frac1x$ is simple, while the polynomial is easy to integrate. (A common memory aid, not in the book, is ILATE: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. L comes before A, so $\log x$ is $u$ and $x^2 + 1$ is $\frac{dv}{dx}$ ⁠.)

Choice of first function: Remark (iii), Section 7.6. Lesson: integration by parts
Apply the integration by parts formula
$\begin{aligned} &\int(x^2+1)\log x\,dx \\ &= \log x\left(\frac{x^3}{3} + x\right) - \int\frac1x\left(\frac{x^3}{3} + x\right)dx. \end{aligned}$
First function times the integral of the second, minus the integral of (derivative of the first) times (integral of the second).

Parts formula. Lesson: integration by parts
Simplify and integrate the leftover polynomial

For $x > 0$ ⁠, $\frac1x\left(\frac{x^3}{3} + x\right) = \frac{x^2}{3} + 1$ ⁠, so
$\int\left(\frac{x^2}{3} + 1\right)dx = \frac{x^3}{9} + x.$
Divide each term by $x$ (allowed, $x \ne 0$ ⁠). Then integrate term by term: $\frac13\cdot\frac{x^3}{3} = \frac{x^3}{9}$ and $\int 1\,dx = x$ ⁠.

Power rule (formula (i)) and linearity, properties (III) and (IV), Section 7.2.1. Basics: powers of x
Write the answer
$\begin{aligned} &\int(x^2+1)\log x\,dx \\ &= \left(\frac{x^3}{3} + x\right)\log x - \frac{x^3}{9} - x + C, \end{aligned}$
for $x > 0$ ⁠.

The minus sign in front of the leftover integral applies to both of its terms. You could also split the question as $\int x^2\log x\,dx + \int\log x\,dx$ (Exercise 7.6 Question 6 and Example 18); adding $\frac{x^3}{3}\log x - \frac{x^3}{9}$ and $x\log x - x$ gives the same answer.

One constant $C$ at the end (Section 7.2). Basics: integration undoes differentiation
Check: differentiate the answer
$\begin{aligned} &\frac{d}{dx}\left[\left(\frac{x^3}{3} + x\right)\log x - \frac{x^3}{9} - x\right] \\ &= (x^2+1)\log x + \left(\frac{x^3}{3} + x\right)\frac1x - \frac{x^2}{3} - 1 \\ &= (x^2+1)\log x + \frac{x^2}{3} + 1 - \frac{x^2}{3} - 1 \\ &= (x^2+1)\log x. \end{aligned}$
The product rule on the first term gives $(x^2 + 1)\log x$ plus a polynomial part $\frac{x^2}{3} + 1$ ⁠, which the derivative of $-\frac{x^3}{9} - x$ cancels exactly. This holds for every $x > 0$ ⁠.

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

Result

\begin{aligned} &\int(x^2+1)\log x\,dx \\ &= \left(\frac{x^3}{3} + x\right)\log x - \frac{x^3}{9} - x + C, \end{aligned}

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

Check it with numbers

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

\begin{aligned} F(e) &= \frac{e^3}{3} + e - \frac{e^3}{9} - e = \frac{2e^3}{9}, \\ F(1) &= 0 - \frac19 - 1 = -\frac{10}{9}, \\ F(e) - F(1) &= \frac{2e^3 + 10}{9} = 5.5745637607\ldots \end{aligned}

Numerical integration of $(x^2+1)\log x$ over $[1, e]$ gives $5.5745637607\ldots$ ⁠, the same.

Common mistakes

Taking the polynomial as the first function. Then $\log x$ would have to be integrated as the second function, which needs another round of parts; possible, but longer and easy to get wrong.
Applying the minus sign to only one term of $\frac{x^3}{9} + x$ ⁠.
Dividing wrongly: $\frac1x\cdot\frac{x^3}{3} = \frac{x^2}{3}$ ⁠, not $\frac{x^2}{3x}$ or $\frac{x^3}{3x^2}$ ⁠.

What to take away

Polynomial times $\log x$ ⁠: take $\log x$ first; the $\frac1x$ turns the integrated polynomial back into a polynomial.
The minus sign in the parts formula applies to the whole leftover integral.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts	Step 2	Section 7.6 (Lesson: integration by parts)
Logarithmic function taken as the first function	Step 1	Remark (iii), Section 7.6
$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ ⁠; integral of a sum	Steps 1, 3	Standard formula (i), Section 7.2; properties (III), (IV), Section 7.2.1
$\int\log x\,dx = x\log x - x + C$	Step 4 (alternative)	Example 18
$\frac{d}{dx}\log x = \frac1x$ ⁠; product rule	Steps 1, 5	Chapter 5, Sections 5.4 and 5.3

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Exercise 7.6 · Q15: ∫(x2+1)log⁡x dx\int(x^2+1)\log x\,dx∫(x2+1)logxdx

Basics you need

Solution

Choose the first and second functions

Apply the integration by parts formula

Simplify and integrate the leftover polynomial