Chapter 7 · Integrals

Exercise 7.7 · Q8: $\int\sqrt{x^2 + 3x}\,dx$

Exercise 7.7Integrals of some more types · Section 7.6.2

The question

Integrate the function

\sqrt{x^2 + 3x}.

Exam answer

Let $I = \displaystyle\int\sqrt{x^2 + 3x}\,dx$ ⁠.

x^2 + 3x = \left(x + \frac32\right)^2 - \left(\frac32\right)^2

Put $x + \frac32 = y$ ⁠. Then $dx = dy$ ⁠.

I = \int\sqrt{y^2 - \left(\frac32\right)^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 - \frac94} \\ &\quad - \frac98\log\left|y + \sqrt{y^2 - \frac94}\right| + C \end{aligned}

$\therefore\ I = \dfrac{2x + 3}{4}\sqrt{x^2 + 3x}$ $- \dfrac98\log\left|x + \dfrac32 + \sqrt{x^2 + 3x}\right|$ $+ C$ ⁠.

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

The one idea this problem is built on

There is no constant term, but completing the square still works and creates one:

x^2 + 3x = \left(x + \frac32\right)^2 - \left(\frac32\right)^2.

That is "square minus $a^2$ " with $a = \frac32$ ⁠, so after $y = x + \frac32$ formula 7.6.2 (i) applies, with the log term subtracted.

Given, to find, and the plan

Given: The integrand $\sqrt{x^2 + 3x}$ ⁠.
To find: $\int\sqrt{x^2 + 3x}\,dx$ ⁠, and where the answer is valid.
Plan: (1) Complete the square. (2) Find the domain. (3) Put $y = x + \frac32$ ⁠. (4) Apply 7.6.2 (i) with $a = \frac32$ ⁠. (5) Return to $x$ ⁠. (6) Differentiate back.

Basics you need

1Completing the square

Half the coefficient of $x$ is $\frac32$ ⁠; its square is $\frac94$ ⁠. Add and subtract $\frac94$ ⁠. 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
$\begin{aligned} x^2 + 3x &= \left(x^2 + 3x + \tfrac94\right) - \tfrac94 \\ &= \left(x + \tfrac32\right)^2 - \left(\tfrac32\right)^2. \end{aligned}$
$\left(x + \frac32\right)^2 = x^2 + 3x + \frac94$ ⁠. We added $\frac94$ ⁠, so we subtract it again. The leftover $-\frac94$ is negative: shape $y^2 - a^2$ with $a = \frac32$ ⁠.

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

We need $x^2 + 3x = x(x + 3) \ge 0$ ⁠:
$x \le -3 \quad\text{or}\quad x \ge 0.$
A product of two factors is $\ge 0$ when both have the same sign (or one is 0): both $\ge 0$ gives $x \ge 0$ ⁠; both $\le 0$ gives $x \le -3$ ⁠. Equivalently $\left|x + \frac32\right| \ge \frac32$ ⁠. For $-3 < x < 0$ the integrand is not real.
Put $y = x + \frac32$

Then $dy = dx$ and
$\int\sqrt{x^2 + 3x}\,dx = \int\sqrt{y^2 - \left(\tfrac32\right)^2}\,dy.$
A shift has derivative 1, so $dx = dy$ ⁠.

Integration by substitution, Section 7.3.1. Basics: substitution
Apply 7.6.2 (i) with $a = \frac32$

Here $a^2 = \frac94$ and $\frac{a^2}{2} = \frac98$ ⁠:
$\begin{aligned} \int\sqrt{y^2 - \tfrac94}\,dy &= \frac{y}{2}\sqrt{y^2 - \tfrac94} \\ &\quad - \frac98\log\left|y + \sqrt{y^2 - \tfrac94}\right| + C. \end{aligned}$
$\frac{9/4}{2} = \frac98$ ⁠. The log term is subtracted because the root is "square minus $a^2$ ".

Formula 7.6.2 (i). Lesson: integration by parts
Return to $x$

With $y = x + \frac32$ ⁠: $\frac{y}{2} = \frac{2x + 3}{4}$ and $y^2 - \frac94 = x^2 + 3x$ ⁠. So
$\begin{aligned} \int\sqrt{x^2 + 3x}\,dx &= \frac{2x + 3}{4}\sqrt{x^2 + 3x} \\ &\quad - \frac98\log\left|x + \tfrac32 + \sqrt{x^2 + 3x}\right| + C. \end{aligned}$
$\frac{x + 3/2}{2} = \frac{2x + 3}{4}$ ⁠. The modulus is needed: for $x \le -3$ ⁠, $x + \frac32 \le -\frac32$ and the root is smaller than $\left|x + \frac32\right|$ ⁠, so the inside of the log is negative there. It is never 0 (that would need $-\frac94 = 0$ after squaring).
Check: differentiate the answer

Write $y = x + \frac32$ ⁠, $r = \sqrt{y^2 - \frac94} = \sqrt{x^2 + 3x}$ ⁠, 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, since $\frac98\cdot\frac1r = \frac{9/4}{2r}$ ⁠,
$\frac{r^2 + y^2 - \frac94}{2r} = \frac{2r^2}{2r} = r = \sqrt{x^2 + 3x}.\ \checkmark$
$\frac r2 = \frac{r^2}{2r}$ and $r^2 = y^2 - \frac94$ ⁠, so the numerator is $2r^2$ ⁠. This holds on both pieces, $x < -3$ and $x > 0$ ⁠; at $x = -3$ and $x = 0$ 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 + 3x}\,dx &= \frac{2x + 3}{4}\sqrt{x^2 + 3x} \\ &\quad - \frac98\log\left|x + \tfrac32 + \sqrt{x^2 + 3x}\right| + C, \end{aligned}

valid for $x \le -3$ and for $x \ge 0$ ⁠. (Textbook answer: $\frac{2x + 3}{4}\sqrt{x^2 + 3x}$ $- \frac98\log\left|x + \frac32 + \sqrt{x^2 + 3x}\right| + C$ ⁠, 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	$F(b) - F(a)$
$[0, 2]$	$3.8566151$	$3.8566151$
$[-6, -4]$	$6.2987898$	$6.2987898$

Common mistakes

"No constant, so no completing the square." The constant appears when you complete the square: $-\frac94$ ⁠.
Coefficient $\frac94$ or $\frac34$ in front of the log. It is $\frac{a^2}{2} = \frac98$ ⁠.
Writing $\log\left|x + \sqrt{x^2 + 3x}\right|$ ⁠. The log contains $y = x + \frac32$ ⁠, not $x$ ⁠.

What to take away

Even $x^2 + bx$ with no constant is completed to $\left(x + \frac b2\right)^2 - \frac{b^2}{4}$ ⁠.
Square minus a positive number: formula 7.6.2 (i), log subtracted, domain in two pieces.

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 $y = x + \frac32$	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 6	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 · Q8: ∫x2+3x dx\int\sqrt{x^2 + 3x}\,dx∫x2+3x​dx

Basics you need

Solution

Complete the square

The domain

Put y=x+32y = x + \frac32y=x+23​

Apply 7.6.2 (i) with a=32a = \frac32a=23​

Return to xxx

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.7 · Q8: $\int\sqrt{x^2 + 3x}\,dx$

Put $y = x + \frac32$

Apply 7.6.2 (i) with $a = \frac32$

Return to $x$