Chapter 7 · Integrals

Exercise 7.2 · Q11: $\int \frac{x}{\sqrt{x+4}}\,dx$

Exercise 7.2Integration by substitution · Section 7.3.1

The question

Integrate the function:

\frac{x}{\sqrt{x+4}},\quad x > 0

Exam answer

Let $I = \displaystyle\int\frac{x}{\sqrt{x + 4}}\,dx$ ⁠.

Put $x + 4 = t$ ⁠. Then $dx = dt$ and $x = t - 4$ ⁠.

\begin{aligned} I &= \int\frac{t - 4}{\sqrt t}\,dt \\ &= \int\left(t^{1/2} - 4t^{-1/2}\right)dt \\ &= \frac{t^{3/2}}{3/2} - 4\cdot\frac{t^{1/2}}{1/2} + C \\ &= \frac23\,t^{3/2} - 8\,t^{1/2} + C \\ &= \frac23\,t^{1/2}(t - 12) + C \end{aligned}

$\therefore\ I = \dfrac23\sqrt{x + 4}\,(x - 8) + C$ ⁠.

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

The one idea this problem is built on

Name the inside of the root $t = x+4$ and write the numerator in terms of $t$ too: $x = t-4$ ⁠. Then $\frac{t-4}{\sqrt t} = t^{1/2} - 4t^{-1/2}$ ⁠, two powers that integrate at once. Taking out a common factor at the end gives the neat form $\frac23\sqrt{x+4}\,(x-8)$ ⁠.

Given, to find, and the plan

Given: $f(x) = \dfrac{x}{\sqrt{x+4}}$ for $x > 0$ ⁠. (The formula itself makes sense whenever $x + 4 > 0$ ⁠, that is $x > -4$ ⁠; the question restricts attention to $x > 0$ ⁠.)
To find: $\displaystyle\int\frac{x}{\sqrt{x+4}}\,dx$ ⁠.
Plan: (1) Substitute $t = x+4$ ⁠, $x = t-4$ ⁠, $dx = dt$ ⁠. (2) Split into powers of $t$ ⁠. (3) Integrate. (4) Return to $x$ and factor. (5) Differentiate to check.

Basics you need

1Substitution, including the leftover $x$

With $t = x+4$ ⁠, every $x$ must be replaced: $x = t-4$ and $dx = dt$ ⁠. Basics: substitution

2Negative and fractional powers

$\frac{1}{\sqrt t} = t^{-1/2}$ and $\frac{t}{\sqrt t} = t^{1/2}$ for $t > 0$ ⁠. Basics: powers of x

3Power formula and linearity

$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$ ⁠, $n \ne -1$ (standard formula (i), Section 7.2); Property (V), Section 7.2.1. Basics: splitting fractions

Solution

Substitute $t = x+4$
$\begin{aligned} t &= x + 4, \\ x &= t - 4, \\ dx &= dt. \end{aligned}$
Therefore
$\int\frac{x}{\sqrt{x+4}}\,dx = \int\frac{t-4}{\sqrt t}\,dt.$
The root is the awkward part, so we name its inside. The numerator $x$ is not the derivative of $x+4$ (that is $1$ ⁠), so instead we express it through $t$ ⁠. For $x > 0$ ⁠, $t > 4$ ⁠, so $\sqrt t$ is defined and non-zero.

Substitution, Section 7.3.1. Lesson: integration by substitution
Split into powers of $t$
$\frac{t-4}{\sqrt t} = \frac{t}{t^{1/2}} - \frac{4}{t^{1/2}} = t^{1/2} - 4t^{-1/2}.$
A fraction whose denominator is a single term splits over the terms of its numerator. Each piece is then a single power of $t$ ⁠.

$\frac{a-b}{c} = \frac ac - \frac bc$ ⁠; $\frac{t^m}{t^k} = t^{m-k}$ ⁠. Basics: splitting fractions
Integrate term by term
$\begin{aligned} \int\left(t^{1/2} - 4t^{-1/2}\right)dt &= \frac{t^{3/2}}{3/2} - 4\cdot\frac{t^{1/2}}{1/2} + C \\ &= \frac23\,t^{3/2} - 8\,t^{1/2} + C. \end{aligned}$
Power formula with $n = \frac12$ (new power $\frac32$ ⁠) and with $n = -\frac12$ (new power $\frac12$ ⁠); neither is $-1$ ⁠, so the formula applies. Dividing by $\frac12$ doubles: $4\cdot 2 = 8$ ⁠.

Standard formula (i), Section 7.2; Property (V), Section 7.2.1.
Return to $x$ and factor

Take out the common factor $\frac23 t^{1/2}$ ⁠, using $t^{3/2} = t\cdot t^{1/2}$ and $8 = \frac23\cdot 12$ ⁠:
$\begin{aligned} \frac23\,t^{3/2} - 8\,t^{1/2} &= \frac23\,t^{1/2}\,(t - 12) \\ &= \frac23\sqrt{x+4}\,(x + 4 - 12) \\ &= \frac23\sqrt{x+4}\,(x - 8). \end{aligned}$
Factoring is not required, but it gives the compact form printed in the answer key and makes the derivative check easier to read.
Check: differentiate the answer

Let $F(x) = \frac23\sqrt{x+4}\,(x-8)$ ⁠. Product rule with $u = \sqrt{x+4}$ ( $u' = \frac{1}{2\sqrt{x+4}}$ ⁠) and $v = x - 8$ ( $v' = 1$ ⁠):
$\begin{aligned} F'(x) &= \frac23\left[\frac{x-8}{2\sqrt{x+4}} + \sqrt{x+4}\right] \\ &= \frac23\cdot\frac{(x-8) + 2(x+4)}{2\sqrt{x+4}} \\ &= \frac23\cdot\frac{3x}{2\sqrt{x+4}} \\ &= \frac{x}{\sqrt{x+4}}. \end{aligned}$
We are back at the integrand.

In the second line both terms are put over the common denominator $2\sqrt{x+4}$ (multiply $\sqrt{x+4}$ by $\frac{2\sqrt{x+4}}{2\sqrt{x+4}}$ ⁠). The numerator collapses to $3x$ ⁠, and $\frac23\cdot\frac32 = 1$ ⁠.

Product rule and chain rule, Chapter 5 (Section 5.3 and Theorem 4). Basics: integration undoes differentiation

Result

For $x > 0$ (in fact for every $x > -4$ ⁠),

\int\frac{x}{\sqrt{x+4}}\,dx = \frac23\sqrt{x+4}\,(x - 8) + C.

(Textbook answer: $\frac23\sqrt{x+4}\,(x-8) + C$ ⁠.) Unfactored, the same answer is $\frac23(x+4)^{3/2} - 8\sqrt{x+4} + C$ ⁠.

Check it with numbers

At $x = 5$ the integrand is $\frac{5}{\sqrt9} = \frac53 \approx 1.666667$ ⁠. With $h = 0.0001$ ⁠:

\begin{aligned} F(5.0001) &\approx -5.9998333321, \\ F(4.9999) &\approx -6.0001666655, \\ \frac{F(5.0001)-F(4.9999)}{0.0002} &\approx 1.666667. \end{aligned}

The slope of $F$ equals the integrand. (The answer itself is negative near $x = 5$ ⁠; that is fine, since only its slope has to match.)

Common mistakes

Treating $x$ as the derivative of $x+4$ ⁠. The derivative of $x+4$ is $1$ ⁠, so $x\,dx$ is not $dt$ ⁠. Replace $x$ by $t-4$ ⁠.
Sign error with the negative power. $\int t^{-1/2}\,dt = 2t^{1/2}$ ⁠, not $-2t^{-3/2}$ (that would be differentiating).
Factoring wrongly. $\frac23 t^{3/2} - 8t^{1/2} = \frac23 t^{1/2}(t - 12)$ ⁠; check by expanding.

What to take away

With a linear inside, write every $x$ through $t$ and split into powers.
Factoring the final answer often matches the textbook's form; expand to confirm the two forms are equal.

Every fact used, and where it comes from

Fact	Used in	Source
Substitution $t = x+4$	Step 1	Section 7.3.1 (Lesson: substitution)
$\frac{a-b}{c} = \frac ac - \frac bc$ ⁠; laws of exponents	Steps 2, 4	Algebra (Basics)
$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$ ⁠, $n \ne -1$	Step 3	Standard formula (i), Section 7.2
Integral of a sum and of a constant multiple	Step 3	Property (V), Section 7.2.1
Product rule; chain rule	Step 5	Chapter 5, Section 5.3 and Theorem 4

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Exercise 7.2 · Q11: ∫xx+4 dx\int \frac{x}{\sqrt{x+4}}\,dx∫x+4​x​dx

Basics you need

Solution

Substitute t=x+4t = x+4t=x+4

Split into powers of ttt

Integrate term by term

Return to xxx and factor

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.2 · Q11: $\int \frac{x}{\sqrt{x+4}}\,dx$

Substitute $t = x+4$

Split into powers of $t$

Return to $x$ and factor