Chapter 7 · Integrals

Exercise 7.2 · Q36: $\int\frac{(x+1)(x+\log x)^2}{x}\,dx$

Exercise 7.2Integration by substitution · Section 7.3.1

The question

Integrate the function (Exercise 7.2, "Integrate the functions in Exercises 1 to 37"):

\frac{(x + 1)(x + \log x)^2}{x}

Exam answer

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

Put $x + \log x = t$ ⁠. Then $\left(1 + \frac1x\right)dx = dt$ ⁠, that is, $\frac{x + 1}{x}\,dx = dt$ ⁠.

I = \int t^2\,dt = \frac{t^3}{3} + C

$\therefore\ I = \dfrac13(x + \log x)^3 + C$ ⁠.

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

The one idea this problem is built on

Differentiate the bracket: $\frac{d}{dx}(x + \log x) = 1 + \frac1x = \frac{x + 1}{x}$ ⁠. That is exactly the rest of the integrand. So with $t = x + \log x$ ⁠, the factor $\frac{x + 1}{x}\,dx$ is $dt$ ⁠, and the integral is $\int t^2\,dt$ ⁠.

Given, to find, and the plan

Given: $f(x) = \dfrac{(x + 1)(x + \log x)^2}{x}$ ⁠, $\log$ the natural logarithm.
To find: $\int f(x)\,dx$ ⁠, and where the answer is valid.
Plan: (1) Domain: $x > 0$ ⁠. (2) Compute the derivative of $x + \log x$ ⁠. (3) Substitute $t = x + \log x$ ⁠. (4) Integrate $t^2$ and return to $x$ ⁠. (5) Differentiate back.

Basics you need

1 $\log x$ and its derivative

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

2Adding fractions

$1 + \frac1x = \frac{x}{x} + \frac1x = \frac{x + 1}{x}$ ⁠. Basics: working with fractions

3Substitution

If $t = g(x)$ ⁠, then $dt = g'(x)\,dx$ ⁠. Basics: substitution (the chain rule backwards)

4Power rule

$\int t^2\,dt = \frac{t^3}{3} + C$ ⁠. Basics: powers and the power rule

Solution

Where is the integrand defined?

$\log x$ needs $x > 0$ ⁠, and then the division by $x$ is allowed. So the integrand is defined for
$x > 0.$
The answer is claimed on this interval (Note, Section 7.2).
Differentiate the bracket $x + \log x$
$\frac{d}{dx}(x + \log x) = 1 + \frac1x = \frac{x + 1}{x}.$
The integrand is a square of $x + \log x$ ⁠, multiplied by $\frac{x + 1}{x}$ ⁠. Computing the bracket's derivative shows that this remaining factor is precisely that derivative. This is the book's advice: substitute for a function whose derivative also occurs in the integrand (Section 7.3.1).

Sum rule for derivatives; $\frac{d}{dx}\log x = \frac1x$ (Section 5.4). Basics: fractions
Substitute $t = x + \log x$

Put $t = x + \log x$ ⁠. By Step 2,
$dt = \frac{x + 1}{x}\,dx.$
Group the integrand to match:
$\begin{aligned} &\int\frac{(x + 1)(x + \log x)^2}{x}\,dx \\ &= \int(x + \log x)^2\cdot\frac{x + 1}{x}\,dx \\ &= \int t^2\,dt. \end{aligned}$
Moving the factor $\frac{x + 1}{x}$ next to $dx$ makes the substitution visible: one part becomes $t^2$ ⁠, the other becomes $dt$ ⁠. Nothing in $x$ is left.

Lesson: integration by substitution
Integrate and return to $x$
$\int t^2\,dt = \frac{t^3}{3} + C = \tfrac13(x + \log x)^3 + C.$
Power rule with $n = 2$ ⁠, then $t = x + \log x$ put back.

Standard formula (i), Section 7.2. Basics: power rule
Check: differentiate the answer back

Let $F(x) = \frac13(x + \log x)^3$ ⁠. Chain rule, outer $\frac13u^3$ ⁠, inner $u = x + \log x$ ⁠:
$\begin{aligned} F'(x) &= \tfrac13\cdot 3(x + \log x)^2\cdot\left(1 + \frac1x\right) \\ &= \frac{(x + 1)(x + \log x)^2}{x}. \end{aligned}$
This is the integrand. ✓

Property (I), Section 7.2.1: differentiating the answer must return the integrand.

Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation

Result

$\int\frac{(x + 1)(x + \log x)^2}{x}\,dx = \tfrac13(x + \log x)^3 + C,$ valid for $x > 0$ ⁠. (Textbook answer: $\frac13(x + \log x)^3 + C$ ⁠.)

Check it with numbers

Take $x = 2$ and $h = 0.001$ ⁠:

$F(2.001) = 6.5220544$ and $F(1.999) = 6.5002952$ ⁠,
slope $\approx \frac{6.5220544 - 6.5002952}{0.002} = 10.8796$ ⁠,
integrand: $\frac{3\,(2 + \log 2)^2}{2} = \frac{3\,(2.693147)^2}{2} = 10.8796$ ⁠. ✓

Common mistakes

Expanding everything. Multiplying out produces terms such as $2x\log x$ and $(\log x)^2$ ⁠, which need integration by parts (Section 7.6). Spotting the derivative avoids all of that.
Taking $t = \log x$ ⁠. Then $dt = \frac{dx}{x}$ leaves $(x + 1)(x + \log x)^2$ with $x = e^t$ inside: much harder.
Forgetting $x > 0$ ⁠.

What to take away

Before expanding, differentiate the most complicated bracket and see if its derivative is the rest of the integrand.
$\frac{d}{dx}(x + \log x) = \frac{x + 1}{x}$ ⁠: combining $1 + \frac1x$ into one fraction is what reveals the match.
Compare Q35: there $t = 1 + \log x$ ⁠, here $t = x + \log x$ ⁠; the method is identical.

Every fact used, and where it comes from

Fact	Used in	Source
$\log x$ defined for $x > 0$ ⁠; $\frac{d}{dx}\log x = \frac1x$	Steps 1, 2	Section 5.4
$1 + \frac1x = \frac{x + 1}{x}$	Step 2	Adding fractions
Substitution $t = x + \log x$	Step 3	Section 7.3.1 (Lesson: substitution)
$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$ ⁠, $n \ne -1$	Step 4	Standard formula (i), Section 7.2
Chain rule; $F' = f$ checks the answer	Step 5	Theorem 4, Section 5.3.1; Property (I), Section 7.2.1

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Exercise 7.2 · Q36: ∫(x+1)(x+log⁡x)2x dx\int\frac{(x+1)(x+\log x)^2}{x}\,dx∫x(x+1)(x+logx)2​dx

Basics you need

Solution

Where is the integrand defined?

Differentiate the bracket x+log⁡xx + \log xx+logx

Substitute t=x+log⁡xt = x + \log xt=x+logx

Integrate and return to xxx

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.2 · Q36: $\int\frac{(x+1)(x+\log x)^2}{x}\,dx$

Differentiate the bracket $x + \log x$

Substitute $t = x + \log x$

Integrate and return to $x$