Chapter 7 · Integrals

Exercise 7.7 · Q11 (MCQ): $\int\sqrt{x^2 - 8x + 7}\,dx$

Exercise 7.7Integrals of some more types · Section 7.6.2

The question

$\int\sqrt{x^2 - 8x + 7}\,dx$ is equal to

(A) $\frac12(x - 4)\sqrt{x^2 - 8x + 7}$ $+ 9\log\left|x - 4 + \sqrt{x^2 - 8x + 7}\right| + C$

(B) $\frac12(x + 4)\sqrt{x^2 - 8x + 7}$ $+ 9\log\left|x + 4 + \sqrt{x^2 - 8x + 7}\right| + C$

(D) $\frac12(x - 4)\sqrt{x^2 - 8x + 7}$ $- \frac92\log\left|x - 4 + \sqrt{x^2 - 8x + 7}\right| + C$

Exam answer

x^2 - 8x + 7 = (x - 4)^2 - 3^2

Put $x - 4 = y$ ⁠. Then $dx = dy$ ⁠.

Using $\displaystyle\int\sqrt{y^2 - a^2}\,dy = \frac{y}{2}\sqrt{y^2 - a^2}$ $- \frac{a^2}{2}\log\left|y + \sqrt{y^2 - a^2}\right| + C$ with ${a = 3,}$

\begin{aligned} &\int\sqrt{y^2 - 3^2}\,dy \\ &= \frac{y}{2}\sqrt{y^2 - 9} \\ &\quad - \frac92\log\left|y + \sqrt{y^2 - 9}\right| + C \end{aligned}

Answer: (D) $\dfrac12(x - 4)\sqrt{x^2 - 8x + 7}$ $- \dfrac92\log\left|x - 4 + \sqrt{x^2 - 8x + 7}\right|$ $+ C$

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

The one idea this problem is built on

Complete the square: $x^2 - 8x + 7 = (x - 4)^2 - 3^2$ ⁠. With $y = x - 4$ this is $\sqrt{y^2 - a^2}$ ⁠, $a = 3$ ⁠, so formula 7.6.2 (i) gives a log term with coefficient $-\frac{a^2}{2} = -\frac92$ ⁠: option (D). Options (A), (C) and (D) differ only in that coefficient, and a short calculation shows that $-\frac92$ is the only one that works.

Given, to find, and the plan

Given: The integrand $\sqrt{x^2 - 8x + 7}$ and four candidate answers.
To find: The correct option.
Plan: (1) Complete the square and find the domain. (2) Apply 7.6.2 (i) with $a = 3$ ⁠. (3) Differentiate back. (4) Examine every option. (5) Conclude.

Basics you need

1Completing the square

$x^2 - 8x = (x - 4)^2 - 16$ ⁠: half of $-8$ is $-4$ ⁠, and $(-4)^2 = 16$ ⁠. Basics: completing the square

2Formula 7.6.2 (i)

For $a > 0$ ⁠, $\int\sqrt{y^2 - a^2}\,dy$ equals $\frac{y}{2}\sqrt{y^2 - a^2} - \frac{a^2}{2}\log\left|y + \sqrt{y^2 - a^2}\right|$ ⁠, plus $C$ (Section 7.6.2; see also Exercise 7.7 Q4). Lesson: integration by parts

3Integration undoes differentiation

An option is correct only if its derivative equals the integrand at every point of the domain. Basics: integration undoes differentiation

Solution

Complete the square and find the domain
$\begin{aligned} x^2 - 8x + 7 &= (x^2 - 8x + 16) + 7 - 16 \\ &= (x - 4)^2 - 9 = (x - 4)^2 - 3^2. \end{aligned}$
The root is real when $|x - 4| \ge 3$ ⁠, that is
$x \le 1 \quad\text{or}\quad x \ge 7.$
$(x - 4)^2 = x^2 - 8x + 16$ ⁠; we added 16, so we subtract it: $7 - 16 = -9$ ⁠. The domain also follows from $x^2 - 8x + 7 = (x - 1)(x - 7) \ge 0$ ⁠. The shape is "square minus $a^2$ " with $a = 3$ ⁠.

Completing the square, Section 7.4 (7). Basics: completing the square
Apply 7.6.2 (i) with $y = x - 4$ ⁠, $a = 3$

Put $y = x - 4$ ⁠, $dy = dx$ ⁠. With $\frac{a^2}{2} = \frac92$ ⁠:
$\begin{aligned} \int\sqrt{x^2 - 8x + 7}\,dx &= \frac12(x - 4)\sqrt{x^2 - 8x + 7} \\ &\quad - \frac92\log\left|x - 4 + \sqrt{x^2 - 8x + 7}\right| + C. \end{aligned}$
$\frac{y}{2} = \frac12(x - 4)$ ⁠, $y^2 - 9 = x^2 - 8x + 7$ ⁠. The log term is subtracted because the root is "square minus $a^2$ ". This is option (D) exactly.

Substitution, Section 7.3.1; formula 7.6.2 (i). Lesson: integration by parts
Check: differentiate the answer

Write $y = x - 4$ ⁠, $r = \sqrt{y^2 - 9}$ ⁠, so $\frac{dr}{dx} = \frac{y}{r}$ where $r > 0$ ⁠, and $\frac{d}{dx}\log|y + r| = \frac{1 + y/r}{y + r} = \frac1r$ ⁠. For any number $k$ ⁠,
$\begin{aligned} &\frac{d}{dx}\left[\frac{y}{2}\,r + k\log|y + r|\right] \\ &= \frac{r}{2} + \frac{y^2}{2r} + \frac{k}{r} = \frac{2y^2 - 9 + 2k}{2r}. \end{aligned}$
With $k = -\frac92$ the numerator is $2y^2 - 18 = 2r^2$ ⁠, so the derivative is $\frac{2r^2}{2r} = r = \sqrt{x^2 - 8x + 7}$ ⁠. $\checkmark$

$\frac r2 = \frac{r^2}{2r}$ and $r^2 = y^2 - 9$ ⁠, so $r^2 + y^2 + 2k = 2y^2 - 9 + 2k$ ⁠. The integrand is $r = \frac{r^2}{r} = \frac{2y^2 - 18}{2r}$ ⁠. Matching numerators: $2k - 9 = -18$ ⁠, so $k = -\frac92$ and no other value. This holds on both pieces of the domain.

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

Examine every option

(A) has $k = 9$ ⁠. By Step 3 its derivative is $\frac{2y^2 + 9}{2r} = r + \frac{27}{2r}$ ⁠, too big by $\frac{27}{2r}$ ⁠. Wrong.
(C) has $k = -3\sqrt2 \approx -4.2426$ ⁠. Its derivative is $r + \frac{9/2 - 3\sqrt2}{r}$ ⁠, and $\frac92 - 3\sqrt2 \approx 0.2574 \ne 0$ ⁠. Wrong, though close.
(B) uses $x + 4$ ⁠, which comes from completing the square with the wrong sign ( $x^2 - 8x$ is $(x - 4)^2 - 16$ ⁠, not $(x + 4)^2 - 16$ ⁠). Its slope at $x = 8$ is about $11.9375916$ ⁠, while the integrand there is $\sqrt7 \approx 2.6457513$ ⁠. Wrong.
(D) has $k = -\frac92$ ⁠: correct by Step 3.

Slopes at $x = 8$ ⁠, where $x^2 - 8x + 7 = 7$ ⁠:

Option	Slope at $x = 8$	Integrand $\sqrt7$	Verdict
(A)	$7.7482717$	$2.6457513$	wrong
(B)	$11.9375916$	$2.6457513$	wrong
(C)	$2.7430240$	$2.6457513$	wrong
(D)	$2.6457513$	$2.6457513$	matches

x = 8

⁠:

y = 4

⁠,

r = \sqrt7

⁠. (A):

\sqrt7 + \frac{27}{2\sqrt7}

⁠. (C):

\sqrt7 + \frac{4.5 - 3\sqrt2}{\sqrt7}

⁠. (B) has a different first term, so Step 3 does not apply to it; its slope was computed directly from the product and chain rules.

Result

\begin{aligned} \int\sqrt{x^2 - 8x + 7}\,dx &= \frac12(x - 4)\sqrt{x^2 - 8x + 7} \\ &\quad - \frac92\log\left|x - 4 + \sqrt{x^2 - 8x + 7}\right| + C, \end{aligned}

valid for $x \le 1$ and for $x \ge 7$ ⁠: option (D). (Textbook answer: D.)

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 (D). One interval is taken on each piece of the domain:

Interval	Sum of thin strips	$F(b) - F(a)$ ⁠, option (D)
$[8, 10]$	$7.9497887$	$7.9497887$
$[-3, 0]$	$13.7141022$	$13.7141022$

Common mistakes

$+$ instead of $-$ before the log. For "square minus $a^2$ " the log term is subtracted; (A) makes this mistake (and uses $a^2$ instead of $\frac{a^2}{2}$ ⁠).
Coefficient $a\sqrt2$ or $a^2$ ⁠. The coefficient is $\frac{a^2}{2} = \frac92$ ⁠; $3\sqrt2$ in (C) is close in value but wrong.
$x + 4$ instead of $x - 4$ ⁠. Half of $-8$ is $-4$ ⁠, so the square is $(x - 4)^2$ ⁠.

What to take away

For $\sqrt{y^2 - a^2}$ the log coefficient is exactly $-\frac{a^2}{2}$ ⁠: differentiating with a general $k$ shows no other value works.
Numbers that are merely close ( $3\sqrt2 \approx 4.243$ against $4.5$ ⁠) are not equal; check exactly.

Every fact used, and where it comes from

Fact	Used in	Source
Completing the square	Step 1	Section 7.4 (7) (Basics: completing the square)
Substitution $y = x - 4$	Step 2	Section 7.3.1
$\int\sqrt{y^2 - a^2}\,dy$ formula	Step 2	Formula 7.6.2 (i), Section 7.6.2
$\frac{d}{dx}\log\|u\| = \frac{u'}{u}$	Steps 3, 4	Standard formula (xii), Section 7.2
$F$ 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 · Q11 (MCQ): ∫x2−8x+7 dx\int\sqrt{x^2 - 8x + 7}\,dx∫x2−8x+7​dx

Basics you need

Solution

Complete the square and find the domain

Apply 7.6.2 (i) with y=x−4y = x - 4y=x−4⁠, a=3a = 3a=3

Check: differentiate the answer

Examine every option

Every fact used, and where it comes from

Exercise 7.7 · Q11 (MCQ): $\int\sqrt{x^2 - 8x + 7}\,dx$

Apply 7.6.2 (i) with $y = x - 4$ ⁠, $a = 3$