Chapter 7 · Integrals

Exercise 7.2 · Q35: $\int\frac{(1+\log x)^2}{x}\,dx$

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

Exam answer

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

Put $1 + \log x = t$ ⁠. Then $\frac{dx}{x} = dt$ ⁠.

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

$\therefore\ I = \dfrac13(1 + \log x)^3 + C$ ⁠.

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

The one idea this problem is built on

The bracket $1 + \log x$ has derivative $\frac1x$ ⁠, and the integrand is divided by $x$ ⁠. So with $t = 1 + \log x$ ⁠, the factor $\frac{dx}{x}$ is exactly $dt$ ⁠, and the integral becomes $\int t^2\,dt$ ⁠.

Given, to find, and the plan

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

Basics you need

1 $\log x$ and its derivative

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

2Substitution

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

3Power rule

$\int t^2\,dt = \frac{t^3}{3} + C$ (standard formula (i), Section 7.2). Basics: powers and the power rule

4Integration undoes differentiation

Check the answer by differentiating it. Basics: integration undoes differentiation

Solution

Where is the integrand defined?

$\log x$ needs $x > 0$ ⁠; then also $x \ne 0$ ⁠, so dividing by $x$ is fine. The integrand is defined exactly for
$x > 0.$
The answer is claimed on this interval (Note, Section 7.2). $1 + \log x$ itself can be negative (for $x < \frac1e$ ⁠); that does no harm, because it is only squared.

Basics: logarithms
Substitute $t = 1 + \log x$

Put $t = 1 + \log x$ ⁠. Differentiate: the constant 1 gives 0 and $\log x$ gives $\frac1x$ ⁠, so
$\frac{dt}{dx} = \frac1x, \quad\text{that is,}\quad dt = \frac{dx}{x}.$
Then
$\int\frac{(1 + \log x)^2}{x}\,dx = \int(1 + \log x)^2\cdot\frac{dx}{x} = \int t^2\,dt.$
The bracket $1 + \log x$ appears raised to a power, and its derivative $\frac1x$ is the remaining factor. That is the signal for substitution (Section 7.3.1). Including the 1 in $t$ means the whole bracket becomes $t$ ⁠.

$\frac{d}{dx}\log x = \frac1x$ ⁠, $x > 0$ (Section 5.4; standard formula (xii), Section 7.2). Lesson: integration by substitution
Integrate and return to $x$
$\int t^2\,dt = \frac{t^3}{3} + C = \tfrac13(1 + \log x)^3 + C.$
Power rule with $n = 2$ ⁠. Then $t$ is replaced by $1 + \log x$ ⁠, because the question is in $x$ ⁠.

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

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

Property (I), Section 7.2.1: $F$ is an antiderivative of $f$ exactly when $F' = f$ ⁠.

Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation
Another way to see it: expand

Expanding the cube,
$\begin{aligned} (1 + \log x)^3 &= 1 + 3\log x \\ &\quad + 3(\log x)^2 + (\log x)^3, \end{aligned}$
so our answer is
$\tfrac13 + \log x + (\log x)^2 + \tfrac13(\log x)^3 + C.$
The $\frac13$ is just a constant, so this is the same family as $\log x + (\log x)^2 + \frac13(\log x)^3 + C$ ⁠, which a student who first expands $(1 + \log x)^2$ and substitutes $t = \log x$ would get.

Both routes are correct; their answers differ only by the constant $\frac13$ (Property (II), Section 7.2.1). Keeping the bracket whole, as in Steps 2–3, is shorter.

Result

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

Check it with numbers

Take $x = 2$ and $h = 0.001$ ⁠:

$F(2.001) \approx 1.6193751837$ and $F(1.999) \approx 1.6165084364$ ⁠,
slope $\approx \frac{1.6193751837 - 1.6165084364}{0.002}$ $\approx 1.43337$ ⁠,
integrand: $\frac{(1 + \log 2)^2}{2} = \frac{(1.693147)^2}{2} = 1.43337$ ⁠. ✓

Common mistakes

Treating $\frac{1}{x}$ as part of the bracket. It is the derivative of $\log x$ ⁠, and becomes $dt$ ⁠.
Integrating the factors separately. The integral of a product is not the product of the integrals: $\frac13(1 + \log x)^3\cdot\log|x|$ is wrong. The factor $\frac1x$ is absorbed into $dt$ ⁠.
Forgetting the domain. The answer holds for $x > 0$ only.

What to take away

Any integrand of the form $\frac{h(\log x)}{x}$ is handled by $t = \log x$ (or $t = a + \log x$ ⁠).
$\int g(x)^n g'(x)\,dx = \frac{g(x)^{n+1}}{n+1} + C$ for $n \ne -1$ ⁠.
Expanding first also works, but the answers then differ by a constant.

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
Substitution $t = 1 + \log x$	Step 2	Section 7.3.1 (Lesson: substitution)
$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$ ⁠, $n \ne -1$	Step 3	Standard formula (i), Section 7.2
Chain rule; $F' = f$ checks the answer	Step 4	Theorem 4, Section 5.3.1; Property (I), Section 7.2.1
Antiderivatives of the same function differ by a constant	Step 5	Property (II), Section 7.2.1

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Exercise 7.2 · Q35: ∫(1+log⁡x)2x dx\int\frac{(1+\log x)^2}{x}\,dx∫x(1+logx)2​dx

Basics you need

Solution

Where is the integrand defined?

Substitute t=1+log⁡xt = 1 + \log xt=1+logx

Integrate and return to xxx

Check: differentiate the answer back

Another way to see it: expand

Every fact used, and where it comes from

Exercise 7.2 · Q35: $\int\frac{(1+\log x)^2}{x}\,dx$

Substitute $t = 1 + \log x$

Integrate and return to $x$