Chapter 7 · Integrals

Exercise 7.2 · Q10: $\int \frac{dx}{x-\sqrt{x}}$

Exercise 7.2Integration by substitution · Section 7.3.1

The question

Integrate the function:

\frac{1}{x-\sqrt{x}}

Exam answer

Let $I = \displaystyle\int\frac{dx}{x - \sqrt x}$ ⁠.

I = \int\frac{dx}{\sqrt x\left(\sqrt x - 1\right)}

Put $\sqrt x - 1 = t$ ⁠. Then $\frac{dx}{2\sqrt x} = dt$ ⁠.

\therefore\ I = \int\frac{2\,dt}{t} = 2\log|t| + C

$\therefore\ I = 2\log\left|\sqrt x - 1\right| + C$ ⁠.

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

The one idea this problem is built on

Since $x = \sqrt x\cdot\sqrt x$ ⁠, the denominator factors as $\sqrt x\,(\sqrt x - 1)$ ⁠. The piece $\frac{1}{\sqrt x}$ is (twice) the derivative of $\sqrt x - 1$ ⁠. So $t = \sqrt x - 1$ turns the integral into $2\int\frac{dt}{t} = 2\log|t|$ ⁠.

Given, to find, and the plan

Given: $f(x) = \dfrac{1}{x - \sqrt x}$ ⁠. It needs $x \ge 0$ (for $\sqrt x$ ⁠) and $x - \sqrt x \ne 0$ ⁠. Since $x - \sqrt x = \sqrt x(\sqrt x - 1)$ ⁠, the denominator vanishes at $x = 0$ and $x = 1$ ⁠. So $f$ is defined for $x > 0$ ⁠, $x \ne 1$ ⁠: on the two intervals $(0, 1)$ and $(1, \infty)$ ⁠.
To find: $\displaystyle\int\frac{dx}{x - \sqrt x}$ ⁠.
Plan: (1) Factor the denominator. (2) Substitute $t = \sqrt x - 1$ ⁠. (3) Integrate $\frac1t$ ⁠. (4) Return to $x$ ⁠. (5) Differentiate to check.

Basics you need

1The square root as a power

$\sqrt x = x^{1/2}$ ⁠, so $\frac{d}{dx}\sqrt x = \frac12 x^{-1/2} = \frac{1}{2\sqrt x}$ for $x > 0$ ⁠. Basics: powers of x

2Substitution

Put $t = g(x)$ ⁠, $dt = g'(x)\,dx$ (Section 7.3.1; Example 5(iii) uses $t = \sqrt x$ in the same way). Basics: substitution

3The integral of $1/t$

$\int\frac{dt}{t} = \log|t| + C$ ⁠, $t \ne 0$ (standard formula (xii), Section 7.2). Basics: logarithms and 1/x

Solution

Factor the denominator

For $x > 0$ ⁠, $x = \left(\sqrt x\right)^2$ ⁠, so
$\begin{aligned} x - \sqrt x &= \sqrt x\left(\sqrt x - 1\right), \\ \frac{1}{x - \sqrt x} &= \frac{1}{\sqrt x - 1}\cdot\frac{1}{\sqrt x}. \end{aligned}$
Factoring reveals a function of $\sqrt x - 1$ multiplied by $\frac{1}{\sqrt x}$ ⁠. The second factor is, up to the constant $2$ ⁠, the derivative of $\sqrt x - 1$ ⁠. That is the substitution pattern.
Substitute $t = \sqrt x - 1$
$\begin{aligned} t &= \sqrt x - 1, \\ dt &= \frac{1}{2\sqrt x}\,dx, \\ \frac{dx}{\sqrt x} &= 2\,dt. \end{aligned}$
Therefore
$\int\frac{1}{\sqrt x - 1}\cdot\frac{dx}{\sqrt x} = \int\frac{1}{t}\cdot 2\,dt = 2\int\frac{dt}{t}.$
We solve $dt = \frac{dx}{2\sqrt x}$ for the combination $\frac{dx}{\sqrt x}$ that is in the integrand. On the domain, $x \ne 1$ means $\sqrt x \ne 1$ ⁠, so $t \ne 0$ and $\frac1t$ is defined.

Substitution, Section 7.3.1; $\frac{d}{dx}\sqrt x = \frac{1}{2\sqrt x}$ ⁠. Lesson: integration by substitution
Integrate in $t$
$2\int\frac{dt}{t} = 2\log|t| + C.$
Standard formula (xii). The absolute value is essential: for $0 < x < 1$ ⁠, $\sqrt x < 1$ and $t$ is negative.

Standard formula (xii), Section 7.2.
Return to $x$
$2\log|t| + C = 2\log\left|\sqrt x - 1\right| + C.$
Replace $t$ by $\sqrt x - 1$ ⁠. The form holds on both intervals $(0, 1)$ and $(1, \infty)$ ⁠.
Check: differentiate the answer

Chain rule with inner $u = \sqrt x - 1$ ( $u' = \frac{1}{2\sqrt x}$ ⁠) and outer $2\log|u|$ (derivative $\frac2u$ ⁠):
$\begin{aligned} \frac{d}{dx}\,2\log\left|\sqrt x - 1\right| &= \frac{2}{\sqrt x - 1}\cdot\frac{1}{2\sqrt x} \\ &= \frac{1}{\sqrt x\left(\sqrt x - 1\right)} \\ &= \frac{1}{x - \sqrt x}. \end{aligned}$
We are back at the integrand, on both intervals of the domain.

The derivative of $\log|u|$ is $\frac1u$ whether $u$ is positive ( $x > 1$ ⁠) or negative ( $0 < x < 1$ ⁠), so this single computation checks the answer on both intervals (Property (I), Section 7.2.1).

$\frac{d}{dx}\log|x| = \frac1x$ ⁠, Section 7.2; chain rule, Chapter 5, Theorem 4. Basics: integration undoes differentiation

Result

For $x > 0$ ⁠, $x \ne 1$ ⁠,

\int\frac{dx}{x - \sqrt x} = 2\log\left|\sqrt x - 1\right| + C.

(Textbook answer: $2\log|\sqrt x - 1| + C$ ⁠.) The constant may differ on $(0, 1)$ and on $(1, \infty)$ ⁠, since the integrand is undefined at $x = 1$ ⁠.

Check it with numbers

Let $F(x) = 2\log|\sqrt x - 1|$ ⁠. At $x = 4$ the integrand is $\frac{1}{4 - 2} = 0.5$ ⁠. With $h = 0.0001$ ⁠:

\begin{aligned} F(4.0001) &\approx 0.0000499991, \\ F(3.9999) &\approx -0.0000500009, \\ \frac{F(4.0001)-F(3.9999)}{0.0002} &\approx 0.500000. \end{aligned}

The slope of $F$ equals the integrand. (Here $F(4) = 2\log 1 = 0$ ⁠, which is why the two values are so close to 0.)

Common mistakes

Splitting the fraction. $\frac{1}{x - \sqrt x}$ is not $\frac1x - \frac{1}{\sqrt x}$ ⁠.
Losing the factor 2. $\frac{dx}{\sqrt x} = 2\,dt$ ⁠, because $dt = \frac{dx}{2\sqrt x}$ ⁠.
Writing $\log(\sqrt x - 1)$ without bars. That is undefined for $0 < x < 1$ ⁠, where the integrand does exist.

What to take away

Treat $\sqrt x$ as a unit: $x = (\sqrt x)^2$ lets you factor expressions in $x$ and $\sqrt x$ ⁠.
A factor $\frac{1}{\sqrt x}$ is a sign that a substitution in $\sqrt x$ will work, since $d(\sqrt x) = \frac{dx}{2\sqrt x}$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$x = (\sqrt x)^2$ for $x \ge 0$	Step 1	Definition of the square root
$\frac{d}{dx}\sqrt x = \frac{1}{2\sqrt x}$ ⁠, $x > 0$	Steps 2, 5	Chapter 5, Table 5.3 ( $n = \frac12$ ⁠); Example 5(iii)
Substitution	Step 2	Section 7.3.1 (Lesson: substitution)
$\int\frac{dt}{t} = \log\|t\| + C$	Step 3	Standard formula (xii), Section 7.2
Chain rule	Step 5	Chapter 5, Theorem 4

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Exercise 7.2 · Q10: ∫dxx−x\int \frac{dx}{x-\sqrt{x}}∫x−x​dx​

Basics you need

Solution

Factor the denominator

Substitute t=x−1t = \sqrt x - 1t=x​−1

Integrate in ttt

Return to xxx

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.2 · Q10: $\int \frac{dx}{x-\sqrt{x}}$

Substitute $t = \sqrt x - 1$

Integrate in $t$

Return to $x$