Chapter 7 · Integrals

Exercise 7.4 · Q15: $\int\frac{dx}{\sqrt{(x - a)(x - b)}}$

Exercise 7.4Integrals of some particular functions · Section 7.4

The question

Integrate the function

\frac{1}{\sqrt{(x - a)(x - b)}}.

Exam answer

Let $I = \displaystyle\int\frac{dx}{\sqrt{(x - a)(x - b)}}$ ⁠.

\begin{aligned} &(x - a)(x - b) \\ &= x^2 - (a + b)x + ab \\ &= \left(x - \frac{a + b}{2}\right)^2 - \left(\frac{a - b}{2}\right)^2 \end{aligned}

Put $x - \frac{a + b}{2} = t$ ⁠. Then $dx = dt$ ⁠.

I = \int\frac{dt}{\sqrt{t^2 - \left(\frac{a - b}{2}\right)^2}}

[Using $\displaystyle\int\frac{dt}{\sqrt{t^2 - k^2}}$ $= \log\left|t + \sqrt{t^2 - k^2}\right| + C$ ]

I = \log\left|t + \sqrt{t^2 - \left(\frac{a - b}{2}\right)^2}\right| + C

$\therefore\ I = \log\Big|x - \dfrac{a + b}{2}$ $+ \sqrt{(x - a)(x - b)}\Big| + C$ (with $a \ne b$ ⁠).

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

The one idea this problem is built on

Multiply out $(x - a)(x - b)$ and complete the square. The result is "a square minus a square":

(x - a)(x - b) = \left(x - \frac{a + b}{2}\right)^2 - \left(\frac{a - b}{2}\right)^2.

That is the shape $\frac{1}{\sqrt{t^2 - k^2}}$ of formula 7.4 (4), whose integral is $\log\left|t + \sqrt{t^2 - k^2}\right|$ ⁠. Here $a$ and $b$ are constants with $a \ne b$ ⁠.

Given, to find, and the plan

Given: The integrand $\frac{1}{\sqrt{(x - a)(x - b)}}$ ⁠, with constants $a \ne b$ ⁠.
To find: $\int\frac{dx}{\sqrt{(x - a)(x - b)}}$ ⁠, and where the answer is valid.
Plan: (1) Find the domain. (2) Expand and complete the square. (3) Shift $t = x - \frac{a + b}{2}$ and use 7.4 (4). (4) Return to $x$ ⁠. (5) Differentiate back, on both pieces of the domain.

Basics you need

1Completing the square

$x^2 - px + q = \left(x - \frac p2\right)^2 + q - \frac{p^2}{4}$ ⁠. Basics: completing the square

2A shift is a substitution

If $t = x - h$ with $h$ a constant, then $dt = dx$ ⁠. Basics: substitution

3Formula 7.4 (4)

$\int\frac{dx}{\sqrt{x^2 - k^2}} = \log\left|x + \sqrt{x^2 - k^2}\right| + C$ ⁠, for $|x| > k > 0$ ⁠. Lesson: integrals of particular functions

4Derivative of $\log|u|$

$\frac{d}{dx}\log|u| = \frac{u'}{u}$ wherever $u \ne 0$ ⁠, used in the check. Basics: logarithms and 1/x

Solution

Where the integrand exists

We need $(x - a)(x - b) > 0$ ⁠: both factors positive or both negative. Writing $m$ and $M$ for the smaller and larger of $a, b$ ⁠, this means
$x < m \quad\text{or}\quad x > M.$
The product is under a square root in a denominator, so it must be strictly positive. Between $m$ and $M$ the two factors have opposite signs. The domain has two separate pieces, and the answer must work on both. (The book does not say so, but we assume $a \ne b$ ⁠: if $a = b$ the integrand would be $\frac{1}{|x - a|}$ ⁠, and the answer below would break down for $x < a$ ⁠, where the quantity inside the logarithm is 0.)
Expand and complete the square
$\begin{aligned} (x - a)(x - b) &= x^2 - (a + b)x + ab \\ &= \left(x - \frac{a + b}{2}\right)^2 - \frac{(a + b)^2}{4} + ab \\ &= \left(x - \frac{a + b}{2}\right)^2 - \frac{(a - b)^2}{4}. \end{aligned}$
Half the coefficient of $x$ is $-\frac{a + b}{2}$ ⁠. The leftover constant is $ab - \frac{(a + b)^2}{4}$ $= \frac{4ab - a^2 - 2ab - b^2}{4}$ $= -\frac{(a - b)^2}{4}$ ⁠, a negative number because $a \ne b$ ⁠.

Type 7.4 (8). Basics: completing the square
Shift and use 7.4 (4)

Put $t = x - \frac{a + b}{2}$ ⁠, so $dt = dx$ ⁠, and let $k = \frac{|a - b|}{2} > 0$ ⁠, so $k^2 = \frac{(a - b)^2}{4}$ ⁠. Then $(x - a)(x - b) = t^2 - k^2$ and
$\begin{aligned} \int\frac{dx}{\sqrt{(x - a)(x - b)}} &= \int\frac{dt}{\sqrt{t^2 - k^2}} \\ &= \log\left|t + \sqrt{t^2 - k^2}\right| + C. \end{aligned}$
"Square minus a positive constant" under a root is the shape of 7.4 (4). On the domain, $t^2 > k^2$ ⁠, that is $|t| > k$ ⁠. The book proves 7.4 (4) with $x = a\sec\theta$ ⁠; the result also holds when $t < -k$ ⁠, because the derivative of $\log\left|t + \sqrt{t^2 - k^2}\right|$ is $\frac{1}{\sqrt{t^2 - k^2}}$ whenever $|t| > k$ (Step 5 shows this). So it covers both pieces of the domain.

Formula 7.4 (4); substitution, Section 7.3.1. Lesson: integrals of particular functions
Return to $x$

Since $t^2 - k^2 = (x - a)(x - b)$ ⁠,
$\begin{aligned} &\int\frac{dx}{\sqrt{(x - a)(x - b)}} \\ &= \log\left|x - \frac{a + b}{2} + \sqrt{(x - a)(x - b)}\right| + C. \end{aligned}$
Undo the shift: $t = x - \frac{a + b}{2}$ ⁠, and by Step 2 the expression under the root is the original product. This is exactly the textbook's form.
Check: differentiate the answer

Write $t = x - \frac{a + b}{2}$ and $s = \sqrt{(x - a)(x - b)} = \sqrt{t^2 - k^2}$ ⁠. Then $s' = \frac{t}{s}$ ⁠, and
$\begin{aligned} \frac{d}{dx}\log|t + s| &= \frac{1 + \frac ts}{t + s} \\ &= \frac{\frac{s + t}{s}}{t + s} = \frac1s. \ \checkmark \end{aligned}$
$s' = \frac{2t}{2s}$ by the chain rule ( $\frac{d}{dx}t^2 = 2t$ because $t' = 1$ ⁠). The sum $t + s$ is never 0: $s^2 = t^2 - k^2 < t^2$ ⁠, so $s < |t|$ ⁠, and $t + s$ has the sign of $t$ ⁠. Nothing here assumed $t > 0$ ⁠, so the check holds on both pieces of the domain. And $\frac1s$ is the integrand.

Chain rule (Theorem 4, Section 5.3.1); derivative (xii), Section 7.2. Basics: logarithms and 1/x

Result

\begin{aligned} &\int\frac{dx}{\sqrt{(x - a)(x - b)}} \\ &= \log\left|x - \frac{a + b}{2} + \sqrt{(x - a)(x - b)}\right| + C, \end{aligned}

valid for $a \ne b$ and every $x$ outside the closed interval between $a$ and $b$ ⁠. (Textbook answer: $\log\left|x - \frac{a + b}{2} + \sqrt{(x - a)(x - b)}\right| + C$ ⁠.)

Check it with numbers

Take $a = 1$ ⁠, $b = 3$ ⁠, so $F(x) = \log\left|x - 2 + \sqrt{(x - 1)(x - 3)}\right|$ and the domain is $x < 1$ or $x > 3$ ⁠. By Section 7.8, $\int_p^q f(x)\,dx = F(q) - F(p)$ ⁠. 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	$F(q) - F(p)$
$[4, 5]$	$0.4457893$	$0.4457893$
$[-1, 0]$	$0.4457893$	$0.4457893$

On $[4, 5]$ ⁠, $F(5) - F(4)$ $= \log\left(3 + 2\sqrt2\right) - \log\left(2 + \sqrt3\right)$ ⁠. The two rows agree because the graph is symmetric about $x = 2$ ⁠.

Common mistakes

Wrong constant after completing the square. $\frac{(a + b)^2}{4} - ab = \frac{(a - b)^2}{4}$ ⁠, not $\frac{(a + b)^2}{4}$ ⁠.
Using the $\sin^{-1}$ formula. The $x^2$ term is positive, so this is "square minus constant": a log, not an inverse sine.
Dropping the modulus. For $x$ below both $a$ and $b$ ⁠, $x - \frac{a + b}{2} + \sqrt{(x - a)(x - b)}$ is negative; the modulus is needed there.

What to take away

$(x - a)(x - b) = \left(x - \frac{a + b}{2}\right)^2 - \left(\frac{a - b}{2}\right)^2$ ⁠: the midpoint of $a$ and $b$ is the centre of the square.
On both pieces of the domain, $\begin{aligned} &\int\frac{dx}{\sqrt{(x - h)^2 - k^2}} \\ &= \log\left|x - h + \sqrt{(x - h)^2 - k^2}\right| + C. \end{aligned}$

Every fact used, and where it comes from

Fact	Used in	Source
Completing the square	Step 2	Type 7.4 (8) (Basics: completing the square)
$\int\frac{dx}{\sqrt{x^2 - a^2}} = \log\left\|x + \sqrt{x^2 - a^2}\right\| + C$	Step 3	Formula 7.4 (4)
Substitution $t = x - \frac{a + b}{2}$	Step 3	Section 7.3.1 (Basics: substitution)
$\frac{d}{dx}\log\|x\| = \frac1x$ ⁠; chain rule	Step 5	Derivative (xii), Section 7.2; Theorem 4, Section 5.3.1
$\int_p^q f(x)\,dx = F(q) - F(p)$	Numerical check	Section 7.8 (Lesson: definite integrals)

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Exercise 7.4 · Q15: ∫dx(x−a)(x−b)\int\frac{dx}{\sqrt{(x - a)(x - b)}}∫(x−a)(x−b)​dx​

Basics you need

Solution

Where the integrand exists

Expand and complete the square

Shift and use 7.4 (4)

Return to xxx

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.4 · Q15: $\int\frac{dx}{\sqrt{(x - a)(x - b)}}$

Return to $x$