Chapter 7 · Integrals

Exercise 7.7 · Q10 (MCQ): $\int\sqrt{1 + x^2}\,dx$

Exercise 7.7Integrals of some more types · Section 7.6.2

The question

$\int\sqrt{1 + x^2}\,dx$ is equal to

(A) $\frac{x}{2}\sqrt{1 + x^2}$ $+ \frac12\log\left|\left(x + \sqrt{1 + x^2}\right)\right| + C$

(B) $\frac23\left(1 + x^2\right)^{\frac32} + C$

(D) $\frac{x^2}{2}\sqrt{1 + x^2}$ $+ \frac12 x^2\log\left|x + \sqrt{1 + x^2}\right| + C$

Exam answer

Using $\displaystyle\int\sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2}$ $+ \frac{a^2}{2}\log\left|x + \sqrt{x^2 + a^2}\right| + C$ with ${a = 1,}$

\begin{aligned} &\int\sqrt{1 + x^2}\,dx \\ &= \frac{x}{2}\sqrt{1 + x^2} \\ &\quad + \frac12\log\left|x + \sqrt{1 + x^2}\right| + C \end{aligned}

Answer: (A) $\dfrac{x}{2}\sqrt{1 + x^2}$ $+ \dfrac12\log\left|x + \sqrt{1 + x^2}\right| + C$

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

The one idea this problem is built on

$\sqrt{1 + x^2} = \sqrt{x^2 + 1^2}$ is formula 7.6.2 (ii) with $a = 1$ ⁠:

\begin{aligned} \int\sqrt{x^2 + a^2}\,dx &= \frac{x}{2}\sqrt{x^2 + a^2} \\ &\quad + \frac{a^2}{2}\log\left|x + \sqrt{x^2 + a^2}\right| + C. \end{aligned}

That is option (A). To be sure the others are wrong, differentiate each: only (A) gives back $\sqrt{1 + x^2}$ ⁠. Even the single point $x = 0$ rules out (B), (C) and (D).

Given, to find, and the plan

Given: The integrand $\sqrt{1 + x^2}$ and four candidate answers.
To find: The correct option.
Plan: (1) Note the domain. (2) Apply 7.6.2 (ii) with $a = 1$ ⁠. (3) Differentiate our answer back. (4) Examine every option by differentiating it. (5) Conclude.

Basics you need

1Integration undoes differentiation

$F$ is an integral of $f$ exactly when $F' = f$ everywhere on the interval. A single point where $F' \ne f$ rules $F$ out. Basics: integration undoes differentiation

2Formula 7.6.2 (ii)

Derived by parts in Exercise 7.7 Q3. Lesson: integration by parts

3Chain rule for powers

$\frac{d}{dx}\left(1 + x^2\right)^{n} = n\left(1 + x^2\right)^{n - 1}\cdot 2x$ ⁠. The factor $2x$ is the derivative of the inside. Basics: substitution and the chain rule

Solution

The domain: every real $x$

$1 + x^2 \ge 1 > 0$ ⁠, so $\sqrt{1 + x^2}$ is defined for all real $x$ ⁠. Also $x + \sqrt{1 + x^2} > 0$ for all $x$ ⁠, because $\sqrt{1 + x^2} > \sqrt{x^2} = |x| \ge -x$ ⁠.

The second fact means the logarithms in (A) and (D) are defined everywhere (the modulus is not even needed).
Apply 7.6.2 (ii) with $a = 1$
$\begin{aligned} \int\sqrt{1 + x^2}\,dx &= \frac{x}{2}\sqrt{1 + x^2} \\ &\quad + \frac12\log\left|x + \sqrt{1 + x^2}\right| + C. \end{aligned}$
$a^2 = 1$ ⁠, so $\frac{a^2}{2} = \frac12$ ⁠. The extra round brackets inside the modulus in option (A) change nothing: this is exactly option (A).

Formula 7.6.2 (ii). Lesson: integration by parts
Check: differentiate the answer

Write $r = \sqrt{1 + x^2}$ ⁠, so $\frac{dr}{dx} = \frac{x}{r}$ ⁠. Then
$\begin{aligned} \frac{d}{dx}\left[\frac{x}{2}\,r\right] &= \frac{r}{2} + \frac{x^2}{2r}, \\ \frac{d}{dx}\left[\frac12\log|x + r|\right] &= \frac12\cdot\frac{1 + \frac{x}{r}}{x + r} = \frac{1}{2r}. \end{aligned}$
Adding, over $2r$ ⁠:
$\begin{aligned} \frac{r^2 + x^2 + 1}{2r} &= \frac{2(1 + x^2)}{2r} \\ &= \frac{r^2}{r} = \sqrt{1 + x^2}.\ \checkmark \end{aligned}$
$1 + \frac{x}{r} = \frac{r + x}{r}$ ⁠, which cancels with $x + r$ ⁠. Then $r^2 = 1 + x^2$ makes the numerator $2r^2$ ⁠. So option (A) really is an integral of $\sqrt{1 + x^2}$ ⁠, for every real $x$ ⁠.

Product rule; chain rule; standard formula (xii), Section 7.2. Basics: logarithms and 1/x

Examine every option

Differentiate each option ( $r = \sqrt{1 + x^2}$ ⁠, $L = \log\left(x + r\right)$ ⁠):

\begin{aligned} \text{(A)}' &= \sqrt{1 + x^2}\quad\text{(Step 3)}, \\ \text{(B)}' &= \frac23\cdot\frac32\,r\cdot 2x = 2x\sqrt{1 + x^2}, \\ \text{(C)}' &= \frac23 r^3 + \frac23 x\cdot\frac32\,r\cdot 2x \\ &= \frac{2 + 8x^2}{3}\sqrt{1 + x^2}, \\ \text{(D)}' &= xr + \frac{x^3 + x^2}{2r} + xL. \end{aligned}

At $x = 0$ the integrand is $\sqrt{1} = 1$ ⁠, and at $x = 1$ it is $\sqrt2 \approx 1.4142136$ ⁠. The slopes of the options there:

Option	Slope at $x = 0$	Slope at $x = 1$	Verdict
(A)	$1$	$1.4142136$	matches
(B)	$0$	$2.8284271$	wrong
(C)	$\frac23$	$4.7140452$	wrong
(D)	$0$	$3.0026939$	wrong

(B) is what you get by treating $\sqrt{1 + x^2}$ like $t^{1/2}$ and raising the power, forgetting that $t = 1 + x^2$ is not $x$ ⁠. It is actually an integral of $2x\sqrt{1 + x^2}$ ⁠, not of $\sqrt{1 + x^2}$ ⁠.
(C) multiplies (B) by $x$ ⁠; its derivative is $\frac{2 + 8x^2}{3}$ times the integrand, which equals the integrand only where $2 + 8x^2 = 3$ ⁠, at two isolated points, not everywhere.
(D) is (A) with the coefficients $\frac{x}{2}$ ⁠, $\frac12$ replaced by $\frac{x^2}{2}$ ⁠, $\frac{x^2}{2}$ ⁠. At $x = 0$ its slope is 0 but the integrand is 1.

An option is correct only if its derivative equals the integrand at every

x

⁠; one point where it fails is enough to reject it. For (B) at 0:

2\cdot 0\cdot 1 = 0

⁠. For (C) at 0:

\frac{2}{3}\cdot 1 = \frac23

⁠. For (D) at 0 every term has a factor

x

⁠, so it is 0. The

x = 1

values:

2\sqrt2

⁠,

\frac{10}{3}\sqrt2

and

\sqrt2 + \frac{1}{\sqrt2} + \log\left(1 + \sqrt2\right)

⁠.

Product rule; chain rule. Basics: substitution and the chain rule

Result

\begin{aligned} \int\sqrt{1 + x^2}\,dx &= \frac{x}{2}\sqrt{1 + x^2} \\ &\quad + \frac12\log\left|x + \sqrt{1 + x^2}\right| + C, \end{aligned}

valid for every real $x$ ⁠: option (A). (Textbook answer: A.)

Check it with numbers

By Section 7.8, $\int_a^b f(x)\,dx = F(b) - F(a)$ for the correct answer $F$ ⁠. The middle column is a computer's sum of very many thin strips, found without any of the options; the last column uses option (A).

Interval	Sum of thin strips	$F(b) - F(a)$ ⁠, option (A)
$[0, 1]$	$1.1477936$	$1.1477936$
$[-1, 2]$	$4.1056793$	$4.1056793$

Common mistakes

Power rule on a function of $1 + x^2$ ⁠. $\int\left(1 + x^2\right)^{1/2}dx \ne \frac23\left(1 + x^2\right)^{3/2}$ ⁠; that would need an extra factor $2x$ in the integrand. This trap gives option (B).
Choosing by appearance. (A) and (D) look alike; only differentiating separates them.
Checking at one point only to confirm. One point can reject an option, but confirming the right one needs the full derivative (Step 3).

What to take away

$\int\sqrt{1 + x^2}\,dx$ equals $\frac{x}{2}\sqrt{1 + x^2}$ plus $\frac12\log\left|x + \sqrt{1 + x^2}\right|$ ⁠, plus $C$ ⁠.
For an integration MCQ, differentiate the options: a quick point such as $x = 0$ often eliminates the wrong ones.

Every fact used, and where it comes from

Fact	Used in	Source
$\int\sqrt{x^2 + a^2}\,dx$ formula	Step 2	Formula 7.6.2 (ii), Section 7.6.2
$\frac{d}{dx}\log\|u\| = \frac{u'}{u}$	Steps 3, 4	Standard formula (xii), Section 7.2
Product rule and chain rule	Steps 3, 4	Differentiation (Basics: the chain rule)
$F$ is an integral of $f$ iff $F' = f$	Step 4	Section 7.2 (Lesson: anti-differentiation)
$\int_a^b f(x)\,dx = F(b) - F(a)$	Numerical check	Section 7.8 (Lesson: definite integrals)

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Exercise 7.7 · Q10 (MCQ): ∫1+x2 dx\int\sqrt{1 + x^2}\,dx∫1+x2​dx

Basics you need

Solution

The domain: every real xxx

Apply 7.6.2 (ii) with a=1a = 1a=1

Check: differentiate the answer

Examine every option

Every fact used, and where it comes from

Exercise 7.7 · Q10 (MCQ): $\int\sqrt{1 + x^2}\,dx$

The domain: every real $x$

Apply 7.6.2 (ii) with $a = 1$