Chapter 7 · Integrals

Exercise 7.7 · Q6: $\int\sqrt{x^2 + 4x - 5}\,dx$

Exercise 7.7Integrals of some more types · Section 7.6.2

The question

Integrate the function

\sqrt{x^2 + 4x - 5}.

Exam answer

Let $I = \displaystyle\int\sqrt{x^2 + 4x - 5}\,dx$ ⁠.

\begin{aligned} x^2 + 4x - 5 &= (x + 2)^2 - 9 \\ &= (x + 2)^2 - 3^2 \end{aligned}

Put $x + 2 = y$ ⁠. Then $dx = dy$ ⁠.

I = \int\sqrt{y^2 - 3^2}\,dy

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

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

$\therefore\ I = \dfrac{x + 2}{2}\sqrt{x^2 + 4x - 5}$ $- \dfrac92\log\left|x + 2 + \sqrt{x^2 + 4x - 5}\right|$ $+ C$ ⁠.

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

The one idea this problem is built on

Completing the square gives $x^2 + 4x - 5 = (x + 2)^2 - 3^2$ ⁠, a square minus a square. After $y = x + 2$ this is $\sqrt{y^2 - a^2}$ with $a = 3$ ⁠, and formula 7.6.2 (i) gives the answer, with the log term subtracted. The leftover is a perfect square here ( $9 = 3^2$ ⁠), so $a$ is a whole number.

Given, to find, and the plan

Given: The integrand $\sqrt{x^2 + 4x - 5}$ ⁠.
To find: $\int\sqrt{x^2 + 4x - 5}\,dx$ ⁠, and where the answer is valid.
Plan: (1) Complete the square. (2) Find the domain. (3) Put $y = x + 2$ ⁠. (4) Apply 7.6.2 (i) with $a = 3$ and return to $x$ ⁠. (5) Differentiate back.

Basics you need

1Completing the square

$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right)$ ⁠. 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 $C$ ⁠; derived by parts in Section 7.6.2 (see also Exercise 7.7 Q4). Lesson: integration by parts

3Derivative of $\log|u|$

$\frac{d}{dx}\log|u| = \frac{u'}{u}$ wherever $u \ne 0$ (standard formula (xii), Section 7.2). Basics: logarithms and 1/x

Solution

Complete the square
$\begin{aligned} x^2 + 4x - 5 &= (x^2 + 4x + 4) - 5 - 4 \\ &= (x + 2)^2 - 9 = (x + 2)^2 - 3^2. \end{aligned}$
Half of 4 is 2 and $2^2 = 4$ ⁠. We added 4 to complete the square, so we subtract it again: $-5 - 4 = -9$ ⁠. The leftover is negative, so the shape is $y^2 - a^2$ with $a = 3$ ⁠.

Completing the square, Section 7.4 (7). Basics: completing the square
The domain

We need $(x + 2)^2 \ge 9$ ⁠, that is $|x + 2| \ge 3$ ⁠:
$x \le -5 \quad\text{or}\quad x \ge 1.$
$x + 2 \ge 3$ gives $x \ge 1$ ⁠; $x + 2 \le -3$ gives $x \le -5$ ⁠. The same comes from factorising: $x^2 + 4x - 5 = (x + 5)(x - 1)$ ⁠, which is $\ge 0$ outside the roots $-5$ and $1$ ⁠. For $-5 < x < 1$ the integrand is not real.
Put $y = x + 2$

Then $dy = dx$ and
$\int\sqrt{x^2 + 4x - 5}\,dx = \int\sqrt{y^2 - 3^2}\,dy.$
A shift has derivative 1, so $dx = dy$ ⁠; the domain becomes $|y| \ge 3$ ⁠.

Integration by substitution, Section 7.3.1. Basics: substitution
Apply 7.6.2 (i) with $a = 3$ and return to $x$
$\begin{aligned} \int\sqrt{y^2 - 9}\,dy &= \frac{y}{2}\sqrt{y^2 - 9} \\ &\quad - \frac92\log\left|y + \sqrt{y^2 - 9}\right| + C. \end{aligned}$
With $y = x + 2$ and $y^2 - 9 = x^2 + 4x - 5$ ⁠:
$\begin{aligned} \int\sqrt{x^2 + 4x - 5}\,dx &= \frac{x + 2}{2}\sqrt{x^2 + 4x - 5} \\ &\quad - \frac92\log\left|x + 2 + \sqrt{x^2 + 4x - 5}\right| + C. \end{aligned}$
$\frac{a^2}{2} = \frac92$ ⁠. The modulus matters: for $x \le -5$ ⁠, $x + 2 \le -3$ and the root is smaller than $|x + 2|$ ⁠, so the expression inside the log is negative. It is never 0, because $\sqrt{(x + 2)^2 - 9} = -(x + 2)$ would give $-9 = 0$ on squaring.

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

Write $y = x + 2$ ⁠, $r = \sqrt{y^2 - 9}$ ⁠, so $\frac{dr}{dx} = \frac{y}{r}$ where $r > 0$ ⁠. Then
$\begin{aligned} \frac{d}{dx}\left[\frac{y}{2}\,r\right] &= \frac{r}{2} + \frac{y^2}{2r}, \\ \frac{d}{dx}\log|y + r| &= \frac{1 + \frac{y}{r}}{y + r} = \frac{1}{r}. \end{aligned}$
So the derivative of our answer is
$\begin{aligned} \frac{r}{2} + \frac{y^2}{2r} - \frac{9}{2r} &= \frac{r^2 + y^2 - 9}{2r} \\ &= \frac{2r^2}{2r} = r.\ \checkmark \end{aligned}$
$r^2 = y^2 - 9$ ⁠, so $r^2 + y^2 - 9 = 2r^2$ ⁠, and $r = \sqrt{x^2 + 4x - 5}$ is the integrand. This works on both pieces of the domain (for $x < -5$ and for $x > 1$ ⁠); at $x = -5$ and $x = 1$ the integrand is 0 and the answer's one-sided slope is 0 too.

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

Result

\begin{aligned} \int\sqrt{x^2 + 4x - 5}\,dx &= \frac{x + 2}{2}\sqrt{x^2 + 4x - 5} \\ &\quad - \frac92\log\left|x + 2 + \sqrt{x^2 + 4x - 5}\right| + C, \end{aligned}

valid for $x \le -5$ and for $x \ge 1$ ⁠. (Textbook answer: $\frac{(x + 2)}{2}\sqrt{x^2 + 4x - 5}$ $- \frac92\log\left|x + 2 + \sqrt{x^2 + 4x - 5}\right| + C$ ⁠, the same.)

Check it with numbers

By Section 7.8, $\int_a^b f(x)\,dx = F(b) - F(a)$ for our answer $F$ ⁠, on any interval where the integrand is continuous. The middle column is a computer's sum of very many thin strips, found without our answer. One interval is taken on each piece of the domain:

Interval	Sum of thin strips	$F(b) - F(a)$
$[1, 3]$	$5.0562447$	$5.0562447$
$[-8, -5]$	$9.6621467$	$9.6621467$

Common mistakes

Leftover $-1$ instead of $-9$ ⁠. After adding 4 we must subtract 4: $-5 - 4 = -9$ ⁠.
Adding the log term. For a square minus $a^2$ ⁠, formula 7.6.2 (i) subtracts $\frac{a^2}{2}\log|\dots|$ ⁠.
Coefficient $\frac32$ instead of $\frac92$ ⁠. The coefficient is $\frac{a^2}{2}$ with $a^2 = 9$ ⁠, not $\frac{a}{2}$ ⁠.

What to take away

Complete the square, read off the sign pattern, then pick 7.6.2 (i), (ii) or (iii).
The domain of $\sqrt{y^2 - a^2}$ is $|y| \ge a$ ⁠: two pieces; the answer works on both.

Every fact used, and where it comes from

Fact	Used in	Source
Completing the square	Steps 1, 2	Section 7.4 (7) (Basics: completing the square)
Substitution $y = x + 2$	Step 3	Section 7.3.1
$\int\sqrt{y^2 - a^2}\,dy$ formula	Step 4	Formula 7.6.2 (i), Section 7.6.2
$\frac{d}{dx}\log\|u\| = \frac{u'}{u}$	Step 5	Standard formula (xii), Section 7.2
$\int_a^b f(x)\,dx = F(b) - F(a)$	Numerical check	Section 7.8 (Lesson: definite integrals)

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Exercise 7.7 · Q6: ∫x2+4x−5 dx\int\sqrt{x^2 + 4x - 5}\,dx∫x2+4x−5​dx

Basics you need

Solution

Complete the square

The domain

Put y=x+2y = x + 2y=x+2

Apply 7.6.2 (i) with a=3a = 3a=3 and return to xxx

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.7 · Q6: $\int\sqrt{x^2 + 4x - 5}\,dx$

Put $y = x + 2$

Apply 7.6.2 (i) with $a = 3$ and return to $x$