Chapter 7 · Integrals

Miscellaneous · Q23: $\int\frac{\sqrt{x^2+1}\,[\log(x^2+1)-2\log x]}{x^4}\,dx$

Miscellaneous ExerciseSubstitution t = 1 + 1/x², then parts with log t · Miscellaneous Exercise

The question

Integrate the function:

\frac{\sqrt{x^2+1}\,\left[\log(x^2+1) - 2\log x\right]}{x^4}

Exam answer

Let $I$ be the given integral. For $x > 0$ ⁠,

\begin{aligned} &\log(x^2 + 1) - 2\log x \\ &= \log\frac{x^2 + 1}{x^2} = \log\left(1 + \frac{1}{x^2}\right) \end{aligned}

\begin{aligned} \frac{\sqrt{x^2 + 1}}{x^4} &= \frac{x\sqrt{1 + \frac{1}{x^2}}}{x^4} \\ &= \frac{1}{x^3}\sqrt{1 + \frac{1}{x^2}} \end{aligned}

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

Put $1 + \frac{1}{x^2} = t$ ⁠. Then $-\frac{2}{x^3}\,dx = dt$ ⁠, so $\frac{dx}{x^3} = -\frac12\,dt$ ⁠.

I = -\frac12\int t^{1/2}\log t\,dt

Taking $\log t$ as the first function and $t^{1/2}$ as the second, by parts:

\begin{aligned} I &= -\frac12\bigg[\frac23t^{3/2}\log t \\ &\qquad - \int\frac1t\cdot\frac23t^{3/2}\,dt\bigg] \\ &= -\frac13t^{3/2}\log t + \frac13\int t^{1/2}\,dt \\ &= -\frac13t^{3/2}\log t + \frac29t^{3/2} + C \\ &= -\frac13t^{3/2}\left[\log t - \frac23\right] + C \end{aligned}

$\therefore\ I = -\dfrac13\left(1 + \dfrac{1}{x^2}\right)^{3/2}$ $\times\left[\log\left(1 + \dfrac{1}{x^2}\right) - \dfrac23\right]$ $+ C$ ⁠.

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

The one idea this problem is built on

Everything in this function is secretly a function of the single quantity $1 + \frac{1}{x^2}$ ⁠: the bracket is $\log\frac{x^2+1}{x^2}$ ⁠, and $\frac{\sqrt{x^2+1}}{x^4} = \frac{1}{x^3}\sqrt{1 + \frac{1}{x^2}}$ ⁠. The leftover $\frac{1}{x^3}$ is (up to $-\frac12$ ⁠) the derivative of $1 + \frac{1}{x^2}$ ⁠. So $t = 1 + \frac{1}{x^2}$ reduces the problem to $\int\sqrt t\log t\,dt$ ⁠, which yields to integration by parts.

Given, to find, and the plan

Given: $f(x) = \frac{\sqrt{x^2+1}\,[\log(x^2+1) - 2\log x]}{x^4}$ ⁠. Because of $\log x$ ⁠, it is defined only for $x > 0$ ⁠.
To find: $\displaystyle\int f(x)\,dx$ ⁠.
Plan: (1) Combine the logarithms. (2) Rewrite the root over $x^4$ ⁠. (3) Substitute $t = 1 + \frac{1}{x^2}$ ⁠. (4) Integrate $\sqrt t\log t$ by parts. (5) Return to $x$ ⁠. (6) Differentiate back and compare with the key.

Basics you need

1Laws of logarithms

For $x > 0$ ⁠: $2\log x = \log x^2$ and $\log A - \log B = \log\frac{A}{B}$ ⁠. Basics: logarithms and 1/x

2Substitution (Section 7.3.1)

$t = g(x)$ ⁠, $dt = g'(x)\,dx$ ⁠. Basics: substitution

3Integration by parts (Section 7.6)

With a logarithm times a power, the logarithm is the first function (Remark (iii), Section 7.6). Lesson: integration by parts

Solution

Combine the logarithms

For $x > 0$ ⁠:
$\begin{aligned}\log(x^2+1) - 2\log x &= \log(x^2+1) - \log x^2\\ &= \log\frac{x^2+1}{x^2} = \log\left(1 + \frac{1}{x^2}\right).\end{aligned}$
The rule $2\log x = \log x^2$ needs $x > 0$ ⁠, which is exactly the domain. Combining the two logarithms into one shows the quantity $1 + \frac{1}{x^2}$ for the first time.

Basics: logarithms and 1/x
Rewrite $\frac{\sqrt{x^2+1}}{x^4}$

For $x > 0$ ⁠,
$\sqrt{x^2+1} = \sqrt{x^2\left(1 + \frac{1}{x^2}\right)} = x\sqrt{1 + \frac{1}{x^2}} ,$
so
$\frac{\sqrt{x^2+1}}{x^4} = \frac{1}{x^3}\sqrt{1 + \frac{1}{x^2}} .$
Hence
$f(x) = \frac{1}{x^3}\sqrt{1 + \frac{1}{x^2}}\;\log\left(1 + \frac{1}{x^2}\right).$
$\sqrt{x^2} = |x| = x$ because $x > 0$ ⁠. Taking the factor $x$ out of the root makes the same quantity $1 + \frac{1}{x^2}$ appear under the root as in the logarithm.
Choose the substitution $t = 1 + \frac{1}{x^2}$
$t = 1 + x^{-2} ,\qquad dt = -2x^{-3}\,dx ,\qquad \frac{dx}{x^3} = -\frac{dt}{2} .$
For $x > 0$ ⁠, $t > 1$ ⁠, and the integral becomes
$\int f(x)\,dx = -\frac12\int\sqrt t\,\log t\,dt .$
Both the root and the logarithm are functions of $1 + \frac{1}{x^2}$ ⁠, and the remaining factor $\frac{1}{x^3}\,dx$ is a constant multiple of its differential. That is precisely when substitution works. Note $t > 1 > 0$ ⁠, so $\sqrt t$ and $\log t$ are defined.

Section 7.3.1. Lesson: integration by substitution
Integrate $\sqrt t\log t$ by parts

First function $\log t$ (derivative $\frac1t$ ⁠), second function $t^{1/2}$ (integral $\frac23t^{3/2}$ ⁠):
$\begin{aligned}\int t^{1/2}\log t\,dt &= \frac23t^{3/2}\log t - \int\frac1t\cdot\frac23t^{3/2}\,dt\\ &= \frac23t^{3/2}\log t - \frac23\int t^{1/2}\,dt\\ &= \frac23t^{3/2}\log t - \frac49t^{3/2} .\end{aligned}$
Multiplying by $-\frac12$ ⁠:
$\begin{aligned}-\frac12\int\sqrt t\log t\,dt &= -\frac13t^{3/2}\log t + \frac29t^{3/2} + C\\ &= -\frac13t^{3/2}\left[\log t - \frac23\right] + C .\end{aligned}$
We cannot integrate $\log t$ directly but can differentiate it, so it is the first function (logarithm before algebraic, Remark (iii)). Its derivative $\frac1t$ lowers the power, and $\frac1t\cdot t^{3/2} = t^{1/2}$ is easy to integrate.

Section 7.6; standard formula (i), Section 7.2. Lesson: integration by parts
Return to $x$
$\begin{aligned}\int f(x)\,dx &= -\frac13\left(1 + \frac{1}{x^2}\right)^{3/2}\\ &\quad\times\left[\log\left(1 + \frac{1}{x^2}\right) - \frac23\right] + C .\end{aligned}$
Replace $t$ by $1 + \frac{1}{x^2}$ ⁠. This is valid for every $x > 0$ ⁠.
Check: differentiate the answer

Write the answer as $F = -\frac13t^{3/2}\left(\log t - \frac23\right)$ with $t = 1 + x^{-2}$ ⁠. By the product rule in $t$ ⁠,
$\begin{aligned}\frac{dF}{dt} &= -\frac13\left[\frac32t^{1/2}\left(\log t - \frac23\right) + t^{3/2}\cdot\frac1t\right]\\ &= -\frac13\left[\frac32t^{1/2}\log t - t^{1/2} + t^{1/2}\right]\\ &= -\frac12\sqrt t\,\log t .\end{aligned}$
Then, by the chain rule with $\frac{dt}{dx} = -\frac{2}{x^3}$ ⁠,
$\frac{dF}{dx} = -\frac12\sqrt t\log t\cdot\left(-\frac{2}{x^3}\right) = \frac{\sqrt t\,\log t}{x^3} = f(x)\ \checkmark$
The last equality is Steps 1 and 2 read backwards. The $-\frac23$ inside the bracket is exactly what cancels the extra $t^{1/2}$ produced by differentiating $\log t$ ⁠.

Chain rule (Chapter 5). Basics: integration undoes differentiation
Compare with the textbook's answer

The key gives $-\frac13\left(1 + \frac{1}{x^2}\right)^{3/2}$ times $\left[\log\left(1 + \frac{1}{x^2}\right) - \frac23\right]$ ⁠, plus $C$ ⁠: exactly our answer.

The key does not state the domain; it is $x > 0$ ⁠, forced by $\log x$ in the question.

Result

For $x > 0$ ⁠:

\begin{aligned}&\int\frac{\sqrt{x^2+1}\,[\log(x^2+1) - 2\log x]}{x^4}\,dx\\ &= -\frac13\left(1 + \frac{1}{x^2}\right)^{3/2}\\ &\quad\times\left[\log\left(1 + \frac{1}{x^2}\right) - \frac23\right] + C\end{aligned}

(Textbook answer: $-\frac13\left(1 + \frac{1}{x^2}\right)^{3/2}$ times $\left[\log\left(1 + \frac{1}{x^2}\right) - \frac23\right]$ ⁠, plus $C$ ⁠; the same as ours.)

Check it with numbers

The rewrite at $x = 1$ ⁠: the original is $\frac{\sqrt2\,(\log 2 - 0)}{1} = \sqrt2\log 2$ ⁠, and $\frac{1}{1}\sqrt{2}\,\log 2$ is the same.

Area from 1 to 2: with $F$ our answer, $F(2) - F(1) \approx 0.2315802003$ ⁠. Numerical integration of the original integrand from 1 to 2 also gives $0.2315802003$ ⁠.

Common mistakes

Expanding instead of combining. Keeping $\log(x^2+1)$ and $2\log x$ apart hides the substitution; combine them into $\log\left(1 + \frac{1}{x^2}\right)$ ⁠.
Taking $x^2$ out of the root as $x^2$ ⁠. $\sqrt{x^2} = x$ (for $x > 0$ ⁠), so only one power of $x$ comes out.
Losing the factor $-\frac12$ ⁠. $dt = -\frac{2}{x^3}\,dx$ ⁠, so $\frac{dx}{x^3} = -\frac{dt}{2}$ ⁠; without it the answer is off by a factor $-2$ ⁠.

What to take away

Look for one quantity that everything depends on; here $1 + \frac{1}{x^2}$ ⁠, with $\frac{dx}{x^3}$ as its differential.
By parts with $\log t$ first, for $n \ne -1$ ⁠: $\int t^n\log t\,dt = \frac{t^{n+1}}{n+1}\log t - \frac{t^{n+1}}{(n+1)^2} + C .$

Every fact used, and where it comes from

Fact	Used in	Source
Laws of logarithms (for $x > 0$ ⁠)	Step 1	Chapter 5, logarithmic functions (Basics)
$\sqrt{x^2} = x$ for $x > 0$	Step 2	Algebra
Integration by substitution	Step 3	Section 7.3.1 (Lesson: substitution)
Integration by parts; logarithm as the first function	Step 4	Section 7.6, Remark (iii) (Lesson: integration by parts)
$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$	Step 4	Standard formula (i), Section 7.2
Product and chain rules	Step 6	Chapter 5

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Miscellaneous · Q23: ∫x2+1 [log⁡(x2+1)−2log⁡x]x4 dx\int\frac{\sqrt{x^2+1}\,[\log(x^2+1)-2\log x]}{x^4}\,dx∫x4x2+1​[log(x2+1)−2logx]​dx

Basics you need

Solution

Combine the logarithms

Rewrite x2+1x4\frac{\sqrt{x^2+1}}{x^4}x4x2+1​​

Choose the substitution t=1+1x2t = 1 + \frac{1}{x^2}t=1+x21​

Integrate tlog⁡t\sqrt t\log tt​logt by parts

Return to xxx

Check: differentiate the answer

Compare with the textbook's answer

Every fact used, and where it comes from

Miscellaneous · Q23: $\int\frac{\sqrt{x^2+1}\,[\log(x^2+1)-2\log x]}{x^4}\,dx$

Rewrite $\frac{\sqrt{x^2+1}}{x^4}$

Choose the substitution $t = 1 + \frac{1}{x^2}$

Integrate $\sqrt t\log t$ by parts

Return to $x$