Chapter 7 · Integrals

Exercise 7.2 · Q6: $\int \sqrt{ax+b}\,dx$

Exercise 7.2Integration by substitution · Section 7.3.1

The question

Integrate the function:

\sqrt{ax+b}

Exam answer

Let $I = \displaystyle\int\sqrt{ax + b}\,dx$ ⁠.

Put $ax + b = t$ ⁠. Then $a\,dx = dt$ ⁠.

\begin{aligned} \therefore\ I &= \frac1a\int t^{1/2}\,dt \\ &= \frac1a\cdot\frac{t^{3/2}}{3/2} + C \end{aligned}

$\therefore\ I = \dfrac{2}{3a}(ax + b)^{3/2} + C$ ( $a \ne 0$ ⁠).

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

The one idea this problem is built on

The square root is applied to a linear expression. Call it $t$ ⁠. Then $dt = a\,dx$ ⁠, so $dx = \frac{dt}{a}$ ⁠, and the integral becomes $\frac1a\int t^{1/2}\,dt$ ⁠, a plain power. The only new feature compared with $\int\sqrt{x}\,dx$ is the factor $\frac1a$ ⁠.

Given, to find, and the plan

Given: $f(x) = \sqrt{ax+b}$ ⁠, with constants $a \ne 0$ and $b$ ⁠. The square root needs $ax + b \ge 0$ ⁠; the integrand is defined (and continuous) exactly there.
To find: $\displaystyle\int\sqrt{ax+b}\,dx$ ⁠.
Plan: (1) Substitute $t = ax+b$ ⁠. (2) Use the power formula with $n = \frac12$ ⁠. (3) Return to $x$ ⁠. (4) Differentiate to check.

Basics you need

1Roots as powers

$\sqrt{t} = t^{1/2}$ for $t \ge 0$ ⁠, and $t^{3/2} = t\sqrt{t}$ ⁠. Basics: powers of x

2Substitution for a linear inside

With $t = ax+b$ ⁠, $dt = a\,dx$ ⁠, so $dx = \frac{dt}{a}$ when $a \ne 0$ ⁠. Basics: substitution

3The power formula

$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$ ⁠, $n \ne -1$ (standard formula (i), Section 7.2). It holds for fractional $n$ too.

Solution

Substitute $t = ax+b$
$\begin{aligned} t &= ax + b, \\ \frac{dt}{dx} &= a, \\ dx &= \frac{dt}{a}. \end{aligned}$
Therefore
$\int\sqrt{ax+b}\,dx = \int\sqrt{t}\cdot\frac{dt}{a} = \frac1a\int t^{1/2}\,dt.$
The square root is the only obstacle, and it sits on a linear expression. Calling that expression $t$ leaves a pure power of $t$ ⁠. Because the inner derivative $a$ is a constant, we can solve for $dx$ and pull $\frac1a$ outside (this is where $a \ne 0$ is needed). On the domain, $t = ax+b \ge 0$ ⁠, so $\sqrt t = t^{1/2}$ makes sense.

Substitution, Section 7.3.1; constant factors move outside, Property (IV), Section 7.2.1. Lesson: integration by substitution
Integrate the power
$\frac1a\int t^{1/2}\,dt = \frac1a\cdot\frac{t^{3/2}}{3/2} + C = \frac{2}{3a}\,t^{3/2} + C.$
The power formula with $n = \frac12$ ⁠: the new power is $\frac12 + 1 = \frac32$ ⁠, and dividing by $\frac32$ is the same as multiplying by $\frac23$ ⁠.

Standard formula (i), Section 7.2. Basics: the power rule
Return to $x$
$\frac{2}{3a}\,t^{3/2} + C = \frac{2}{3a}(ax+b)^{3/2} + C.$
Replace $t$ by $ax+b$ ⁠. The answer is defined wherever the integrand is, namely $ax + b \ge 0$ ⁠.
Check: differentiate the answer

Chain rule with inner $u = ax+b$ ( $u' = a$ ⁠) and outer $\frac{2}{3a}u^{3/2}$ ⁠, for $ax+b > 0$ ⁠:
$\begin{aligned} \frac{d}{dx}\left[\frac{2}{3a}(ax+b)^{3/2}\right] &= \frac{2}{3a}\cdot\frac32(ax+b)^{1/2}\cdot a \\ &= (ax+b)^{1/2} \\ &= \sqrt{ax+b}. \end{aligned}$
The constants $\frac{2}{3a}$ ⁠, $\frac32$ and $a$ multiply to $1$ ⁠, and we are back at the integrand.

This confirms the answer (Property (I), Section 7.2.1). At the single point where $ax+b = 0$ the formula $\frac32 u^{1/2}$ still gives $0$ as a one-sided derivative, matching $\sqrt0 = 0$ ⁠.

Chain rule, Chapter 5, Theorem 4; $\frac{d}{du}u^{n} = nu^{n-1}$ ⁠. Basics: integration undoes differentiation

Result

For $a \ne 0$ ⁠, wherever $ax + b \ge 0$ ⁠,

\int\sqrt{ax+b}\,dx = \frac{2}{3a}(ax+b)^{3/2} + C.

(Textbook answer: $\frac{2}{3a}(ax+b)^{3/2} + C$ ⁠.)

Check it with numbers

Take $a = 3$ ⁠, $b = 1$ ⁠, so $F(x) = \frac29(3x+1)^{3/2}$ ⁠. At $x = 1$ the integrand is $\sqrt{4} = 2$ ⁠. With $h = 0.0001$ ⁠:

\begin{aligned} F(1.0001) &\approx 1.77797778, \\ F(0.9999) &\approx 1.77757778, \\ \frac{F(1.0001)-F(0.9999)}{0.0002} &\approx 2.000000. \end{aligned}

The slope of $F$ equals the integrand.

Common mistakes

Forgetting the $\frac1a$ ⁠. $\frac23(ax+b)^{3/2}$ differentiates to $a\sqrt{ax+b}$ ⁠, not $\sqrt{ax+b}$ ⁠.
Dividing by $\frac12$ instead of $\frac32$ ⁠. Raise the power first, then divide by the new power.
Ignoring the domain. The answer only makes sense where $ax+b \ge 0$ ⁠.

What to take away

For a linear inside $ax+b$ with $a \ne 0$ ⁠: integrate as if the inside were $x$ ⁠, then divide by $a$ ⁠.
So, for $n \ne -1$ ⁠, $\int(ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C.$

Every fact used, and where it comes from

Fact	Used in	Source
$\sqrt t = t^{1/2}$ ⁠, $t \ge 0$	Step 1	Laws of exponents (Basics)
Substitution with a linear inner function	Step 1	Section 7.3.1 (Lesson: substitution)
$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$ ⁠, $n \ne -1$	Step 2	Standard formula (i), Section 7.2
Constant factors move outside	Step 1	Property (IV), Section 7.2.1
Chain rule; power rule	Step 4	Chapter 5, Theorem 4 and Table 5.3

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Exercise 7.2 · Q6: ∫ax+b dx\int \sqrt{ax+b}\,dx∫ax+b​dx

Basics you need

Solution

Substitute t=ax+bt = ax+bt=ax+b

Integrate the power

Return to xxx

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.2 · Q6: $\int \sqrt{ax+b}\,dx$

Substitute $t = ax+b$

Return to $x$