Chapter 7 · Integrals

Exercise 7.2 · Q7: $\int x\sqrt{x+2}\,dx$

Exercise 7.2Integration by substitution · Section 7.3.1

The question

Integrate the function:

x\sqrt{x+2}

Exam answer

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

Put $x + 2 = t$ ⁠. Then $dx = dt$ and $x = t - 2$ ⁠.

\begin{aligned} \therefore\ I &= \int(t - 2)\sqrt{t}\,dt \\ &= \int\left(t^{3/2} - 2t^{1/2}\right)dt \\ &= \frac{t^{5/2}}{5/2} - 2\cdot\frac{t^{3/2}}{3/2} + C \end{aligned}

$\therefore\ I = \dfrac25(x + 2)^{5/2}$ $- \dfrac43(x + 2)^{3/2} + C$ ⁠.

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

The one idea this problem is built on

Put $t = x+2$ to tidy the square root. The factor $x$ outside is not the derivative of anything useful, but it can be rewritten in terms of $t$ : $x = t-2$ ⁠. Then the integrand is $(t-2)\sqrt t = t^{3/2} - 2t^{1/2}$ ⁠, two plain powers.

Given, to find, and the plan

Given: $f(x) = x\sqrt{x+2}$ ⁠, defined and continuous for $x \ge -2$ ⁠.
To find: $\displaystyle\int x\sqrt{x+2}\,dx$ ⁠.
Plan: (1) Substitute $t = x+2$ ⁠, so $x = t-2$ and $dx = dt$ ⁠. (2) Expand into powers of $t$ ⁠. (3) Integrate term by term. (4) Return to $x$ ⁠. (5) Differentiate to check.

Basics you need

1Substitution, including the leftover $x$

After choosing $t = g(x)$ ⁠, every remaining $x$ must also be written in terms of $t$ ⁠. With $t = x+2$ ⁠, that means $x = t-2$ ⁠. Basics: substitution

2Laws of exponents

$t\cdot t^{1/2} = t^{3/2}$ and $t^{5/2} = t^{3/2}\cdot t$ ⁠. 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); integrals of sums and constant multiples split up (Property (V), Section 7.2.1).

Solution

Substitute $t = x+2$
$\begin{aligned} t &= x + 2, \\ x &= t - 2, \\ dx &= dt. \end{aligned}$
Therefore
$\int x\sqrt{x+2}\,dx = \int (t-2)\sqrt{t}\,dt.$
The square root of $x+2$ is the awkward part, so we name its inside $t$ ⁠. Since $\frac{dt}{dx} = 1$ ⁠, $dx = dt$ ⁠. The factor $x$ is not $dt$ -material, so we express it through $t$ ⁠: from $t = x+2$ we get $x = t-2$ ⁠. Now no $x$ remains. On the domain $x \ge -2$ we have $t \ge 0$ ⁠, so $\sqrt t$ is defined.

Substitution, Section 7.3.1. Lesson: integration by substitution
Expand into powers of $t$
$(t-2)\sqrt t = t\cdot t^{1/2} - 2t^{1/2} = t^{3/2} - 2t^{1/2}.$
The power formula only handles single powers. Multiplying out turns a product into a sum of powers, which can be integrated term by term.

$t^{m}\cdot t^{k} = t^{m+k}$ ⁠. Basics: powers of x
Integrate term by term
$\begin{aligned} \int\left(t^{3/2} - 2t^{1/2}\right)dt &= \frac{t^{5/2}}{5/2} - 2\cdot\frac{t^{3/2}}{3/2} + C \\ &= \frac25\,t^{5/2} - \frac43\,t^{3/2} + C. \end{aligned}$
The power formula with $n = \frac32$ (new power $\frac52$ ⁠) and with $n = \frac12$ (new power $\frac32$ ⁠). Dividing by $\frac52$ means multiplying by $\frac25$ ⁠; $2$ divided by $\frac32$ is $\frac43$ ⁠.

Standard formula (i), Section 7.2; Property (V), Section 7.2.1.
Return to $x$
$\frac25\,t^{5/2} - \frac43\,t^{3/2} + C = \frac25(x+2)^{5/2} - \frac43(x+2)^{3/2} + C.$
Replace $t$ by $x+2$ ⁠. This is exactly the textbook's form.
Check: differentiate the answer

Let $F(x) = \frac25(x+2)^{5/2} - \frac43(x+2)^{3/2}$ ⁠. For $x > -2$ ⁠, the chain rule with inner $x+2$ (derivative $1$ ⁠) gives
$\begin{aligned} F'(x) &= \tfrac25\cdot\tfrac52(x+2)^{3/2} - \tfrac43\cdot\tfrac32(x+2)^{1/2} \\ &= (x+2)^{3/2} - 2(x+2)^{1/2} \\ &= (x+2)^{1/2}\big[(x+2) - 2\big] \\ &= x\sqrt{x+2}. \end{aligned}$
The coefficients work out because $\frac25\cdot\frac52 = 1$ and $\frac43\cdot\frac32 = 2$ ⁠. Taking out the common factor $(x+2)^{1/2}$ returns the integrand.

Getting the integrand back confirms the antiderivative (Property (I), Section 7.2.1). The factoring step is the reverse of Step 2.

Chain rule, Chapter 5, Theorem 4. Basics: integration undoes differentiation

Result

For $x \ge -2$ ⁠,

\int x\sqrt{x+2}\,dx = \frac25(x+2)^{5/2} - \frac43(x+2)^{3/2} + C.

(Textbook answer: $\frac25(x+2)^{5/2} - \frac43(x+2)^{3/2} + C$ ⁠.) Factored, this is also $\frac{2}{15}(x+2)^{3/2}(3x-4) + C$ ⁠.

Check it with numbers

Let $F(x) = \frac25(x+2)^{5/2} - \frac43(x+2)^{3/2}$ ⁠. At $x = 2$ the integrand is $2\sqrt4 = 4$ ⁠. With $h = 0.0001$ ⁠:

\begin{aligned} F(2.0001) &\approx 2.13373335, \\ F(1.9999) &\approx 2.13293335, \\ \frac{F(2.0001)-F(1.9999)}{0.0002} &\approx 4.000000. \end{aligned}

The slope of $F$ equals the integrand.

Common mistakes

Leaving an $x$ inside a $dt$ -integral. $\int x\sqrt t\,dt$ is meaningless as it stands; $x$ must become $t-2$ ⁠.
Integrating the two factors separately. $\int x\sqrt{x+2}\,dx$ is not $\frac{x^2}{2}\cdot\frac23(x+2)^{3/2}$ ⁠; there is no product rule for integrals.
Sign slip in $x = t-2$ ⁠. From $t = x+2$ ⁠, $x = t-2$ ⁠, not $t+2$ ⁠.

What to take away

If a leftover factor is not the derivative of the inner function, try expressing it through $t$ and expanding.
This works whenever the inner function is linear, since then $x$ is easy to solve for.

Every fact used, and where it comes from

Fact	Used in	Source
Substitution $t = x+2$ ⁠, $dx = dt$	Step 1	Section 7.3.1 (Lesson: substitution)
$t^m t^k = t^{m+k}$	Steps 2, 5	Laws of exponents
$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$ ⁠, $n \ne -1$	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
Chain rule	Step 5	Chapter 5, Theorem 4

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Exercise 7.2 · Q7: ∫xx+2 dx\int x\sqrt{x+2}\,dx∫xx+2​dx

Basics you need

Solution

Substitute t=x+2t = x+2t=x+2

Expand into powers of ttt

Integrate term by term

Return to xxx

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.2 · Q7: $\int x\sqrt{x+2}\,dx$

Substitute $t = x+2$

Expand into powers of $t$

Return to $x$