Chapter 7 · Integrals

Exercise 7.4 · Q17: $\int\frac{x + 2}{\sqrt{x^2 - 1}}\,dx$

Exercise 7.4Integrals of some particular functions · Section 7.4

The question

Integrate the function

\frac{x + 2}{\sqrt{x^2 - 1}}.

Exam answer

Let $I = \displaystyle\int\frac{x + 2}{\sqrt{x^2 - 1}}\,dx$ ⁠.

Let $x + 2 = A(2x) + B$ ⁠, where $2x = \frac{d}{dx}(x^2 - 1)$ ⁠.

Comparing coefficients: $2A = 1$ and $B = 2$ ⁠. So $A = \frac12$ ⁠, $B = 2$ ⁠.

\begin{aligned} \therefore\ I &= \frac12\int\frac{2x\,dx}{\sqrt{x^2 - 1}} \\ &\quad + 2\int\frac{dx}{\sqrt{x^2 - 1}} \\ &= \frac12 I_1 + 2I_2 \end{aligned}

In $I_1$ ⁠, put $x^2 - 1 = t$ ⁠. Then $2x\,dx = dt$ ⁠.

\begin{aligned} I_1 &= \int\frac{dt}{\sqrt t} = 2\sqrt t \\ &= 2\sqrt{x^2 - 1} \end{aligned}

[Using $\displaystyle\int\frac{dx}{\sqrt{x^2 - a^2}}$ $= \log\left|x + \sqrt{x^2 - a^2}\right| + C$ ]

\begin{aligned} I_2 &= \int\frac{dx}{\sqrt{x^2 - 1^2}} \\ &= \log\left|x + \sqrt{x^2 - 1}\right| \end{aligned}

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

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

The one idea this problem is built on

Split the numerator as a multiple of the derivative of $Q = x^2 - 1$ ⁠, plus a constant: $x + 2 = \frac12(2x) + 2$ (type 7.4 (10)). The $2x$ part is a perfect substitution $t = x^2 - 1$ ⁠; the constant part leaves $2\int\frac{dx}{\sqrt{x^2 - 1}}$ ⁠, which is formula 7.4 (4) with $a = 1$ ⁠.

Given, to find, and the plan

Given: The integrand $\frac{x + 2}{\sqrt{x^2 - 1}}$ ⁠.
To find: $\int\frac{x + 2}{\sqrt{x^2 - 1}}\,dx$ ⁠, and where the answer is valid.
Plan: (1) Find the domain. (2) Find $A$ ⁠, $B$ with $x + 2 = A(2x) + B$ ⁠. (3) Split into $I_1$ and $I_2$ ⁠. (4) $I_1$ by substitution. (5) $I_2$ by 7.4 (4). (6) Combine. (7) Differentiate back.

Basics you need

1Comparing coefficients

If $px + q = A(2ax + b) + B$ for all $x$ ⁠, the coefficients of $x$ and the constants on the two sides are equal. Basics: splitting fractions

2Substitution $t = Q(x)$

$\int\frac{Q'(x)}{\sqrt{Q(x)}}\,dx = 2\sqrt{Q(x)} + C$ ⁠. Basics: substitution

3Formula 7.4 (4)

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

Solution

Where the integrand exists

We need $x^2 - 1 > 0$ ⁠, that is $x < -1$ or $x > 1$ ⁠.

The quadratic is under a square root in a denominator, so it must be strictly positive. The domain has two separate pieces; the answer must work on both.
Split the numerator as $A\cdot Q' + B$

$Q = x^2 - 1$ has $Q' = 2x$ ⁠. Write $x + 2 = A(2x) + B$ ⁠.

Coefficients of $x$ ⁠: $2A = 1$ ⁠. Constants: $B = 2$ ⁠. So $A = \frac12$ ⁠, $B = 2$ ⁠.

This is type 7.4 (10). Check: $\frac12(2x) + 2 = x + 2$ ⁠.

Type 7.4 (10). Basics: splitting fractions
Split the integral into $I_1$ and $I_2$
$\begin{aligned} \int\frac{x + 2}{\sqrt{x^2 - 1}}\,dx &= \frac12\int\frac{2x\,dx}{\sqrt{x^2 - 1}} + 2\int\frac{dx}{\sqrt{x^2 - 1}} \\ &= \frac12 I_1 + 2I_2. \end{aligned}$
Dividing $\frac12(2x) + 2$ by the denominator term by term gives two fractions, and Property (V) of Section 7.2.1 splits the integral of a sum of constant multiples.

Property (V), Section 7.2.1. Lesson: integrals of particular functions
$I_1$ by substitution

Put $t = x^2 - 1$ (so $t > 0$ ⁠), $dt = 2x\,dx$ ⁠:
$\begin{aligned} I_1 = \int\frac{dt}{\sqrt t} &= 2\sqrt t + C_1 \\ &= 2\sqrt{x^2 - 1} + C_1. \end{aligned}$
Power formula with $n = -\frac12$ ⁠: $\int t^{-1/2}\,dt = \frac{t^{1/2}}{1/2} = 2\sqrt t$ ⁠.

Substitution, Section 7.3.1; standard formula (i), Section 7.2. Basics: the power rule
$I_2$ by formula 7.4 (4)
$I_2 = \int\frac{dx}{\sqrt{x^2 - 1^2}} = \log\left|x + \sqrt{x^2 - 1}\right| + C_2.$
"Square minus a positive constant" under a root is the shape of 7.4 (4), here with $a = 1$ ⁠. The book proves it with $x = a\sec\theta$ ⁠; it also holds for $x < -1$ ⁠, because the derivative of $\log\left|x + \sqrt{x^2 - 1}\right|$ is $\frac{1}{\sqrt{x^2 - 1}}$ whenever $|x| > 1$ (shown in Step 7).

Formula 7.4 (4). Lesson: integrals of particular functions
Combine
$\begin{aligned} \int\frac{x + 2}{\sqrt{x^2 - 1}}\,dx &= \sqrt{x^2 - 1} \\ &\quad + 2\log\left|x + \sqrt{x^2 - 1}\right| + C, \end{aligned}$
with $C = \frac{C_1}{2} + 2C_2$ ⁠.

$\frac12\cdot 2\sqrt{x^2 - 1} = \sqrt{x^2 - 1}$ ⁠; the two arbitrary constants merge into one. This is exactly the textbook's form.
Check: differentiate the answer

Write $s = \sqrt{x^2 - 1}$ ⁠, so $s' = \frac xs$ ⁠. Then
$\begin{aligned} \frac{d}{dx}\log|x + s| &= \frac{1 + \frac xs}{x + s} = \frac{\frac{s + x}{s}}{x + s} = \frac1s, \\ \frac{d}{dx}\left(s + 2\log|x + s|\right) &= \frac xs + \frac2s = \frac{x + 2}{\sqrt{x^2 - 1}}. \ \checkmark \end{aligned}$
$s' = \frac{2x}{2s}$ by the chain rule. The sum $x + s$ is never 0: $s^2 = x^2 - 1 < x^2$ ⁠, so $s < |x|$ and $x + s$ has the sign of $x$ ⁠. Nothing here assumed $x > 1$ ⁠, so the check holds on both pieces of the domain.

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

Result

\begin{aligned} \int\frac{x + 2}{\sqrt{x^2 - 1}}\,dx &= \sqrt{x^2 - 1} \\ &\quad + 2\log\left|x + \sqrt{x^2 - 1}\right| + C, \end{aligned}

valid for $x < -1$ or $x > 1$ ⁠. (Textbook answer: $\sqrt{x^2 - 1} + 2\log\left|x + \sqrt{x^2 - 1}\right| + C$ ⁠.)

Check it with numbers

By Section 7.8, $\int_a^b f(x)\,dx = F(b) - F(a)$ for our answer $F$ ⁠. 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	$F(b) - F(a)$
$[2, 3]$	$1.9879549$	$1.9879549$
$[-3, -2]$	$-0.2047978$	$-0.2047978$

On $[2, 3]$ ⁠, $F(3) - F(2) = 2\sqrt2 - \sqrt3 + 2\log\frac{3 + 2\sqrt2}{2 + \sqrt3}$ ⁠.

Common mistakes

Taking $\int\frac{x\,dx}{\sqrt{x^2 - 1}}$ as a log. The numerator $x$ is half the derivative of $x^2 - 1$ ⁠, and over a square root the result is $\sqrt{x^2 - 1}$ ⁠, not a log.
Using $\sin^{-1}x$ for $\int\frac{dx}{\sqrt{x^2 - 1}}$ ⁠. $\sin^{-1}$ belongs to $\frac{1}{\sqrt{1 - x^2}}$ ⁠; here the square comes first, so it is the log formula.
Forgetting the factor 2 in front of $I_2$ ⁠. The constant $B = 2$ multiplies the whole log term.

What to take away

Linear over the root of a quadratic: numerator $= A\cdot Q' + B$ ⁠, then $2A\sqrt Q$ plus $B$ times a standard root integral.
$\int\frac{dx}{\sqrt{x^2 - 1}} = \log\left|x + \sqrt{x^2 - 1}\right| + C$ on both $x > 1$ and $x < -1$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$px + q = A\frac{d}{dx}(ax^2 + bx + c) + B$	Step 2	Type 7.4 (10) (Lesson: integrals of particular functions)
Integral of a sum of constant multiples	Step 3	Property (V), Section 7.2.1
$\int t^{-1/2}\,dt = 2\sqrt t + C$	Step 4	Standard formula (i), Section 7.2
Substitution $t = x^2 - 1$	Step 4	Section 7.3.1 (Basics: substitution)
$\int\frac{dx}{\sqrt{x^2 - a^2}} = \log\left\|x + \sqrt{x^2 - a^2}\right\| + C$	Step 5	Formula 7.4 (4)
$\frac{d}{dx}\log\|x\| = \frac1x$ ⁠; chain rule	Step 7	Derivative (xii), Section 7.2; Theorem 4, Section 5.3.1
$\int_a^b f(x)\,dx = F(b) - F(a)$	Numerical check	Section 7.8 (Lesson: definite integrals)

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Exercise 7.4 · Q17: ∫x+2x2−1 dx\int\frac{x + 2}{\sqrt{x^2 - 1}}\,dx∫x2−1​x+2​dx

Basics you need

Solution

Where the integrand exists

Split the numerator as A⋅Q′+BA\cdot Q' + BA⋅Q′+B

Split the integral into I1I_1I1​ and I2I_2I2​

I1I_1I1​ by substitution

I2I_2I2​ by formula 7.4 (4)

Combine

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.4 · Q17: $\int\frac{x + 2}{\sqrt{x^2 - 1}}\,dx$

Split the numerator as $A\cdot Q' + B$

Split the integral into $I_1$ and $I_2$

$I_1$ by substitution

$I_2$ by formula 7.4 (4)