Chapter 7 · Integrals

Example 10: $\int\frac{x + 2}{2x^2 + 6x + 5}\,dx$ and $\int\frac{x + 3}{\sqrt{5 - 4x - x^2}}\,dx$

ExamplesIntegrals of some particular functions · Section 7.4

The question

Find the following integrals:

(i) $\displaystyle\int\frac{x + 2}{2x^2 + 6x + 5}\,dx$ (ii) $\displaystyle\int\frac{x + 3}{\sqrt{5 - 4x - x^2}}\,dx$

Exam answer

(i) Let $I = \displaystyle\int\frac{x + 2}{2x^2 + 6x + 5}\,dx$ ⁠.

Let $x + 2 = A(4x + 6) + B$ ⁠, where $4x + 6 = \frac{d}{dx}(2x^2 + 6x + 5)$ ⁠.

Comparing coefficients: $4A = 1$ and $6A + B = 2$ ⁠. So $A = \frac14$ ⁠, $B = \frac12$ ⁠.

\begin{aligned} \therefore\ I &= \frac14\int\frac{(4x + 6)\,dx}{2x^2 + 6x + 5} \\ &\quad + \frac12\int\frac{dx}{2x^2 + 6x + 5} \\ &= \frac14 I_1 + \frac12 I_2 \end{aligned}

In $I_1$ ⁠, put $2x^2 + 6x + 5 = t$ ⁠. Then $(4x + 6)\,dx = dt$ ⁠.

\begin{aligned} I_1 &= \int\frac{dt}{t} = \log|t| \\ &= \log|2x^2 + 6x + 5| \end{aligned}

For $I_2$ ⁠, complete the square:

\begin{aligned} I_2 &= \frac12\int\frac{dx}{x^2 + 3x + \frac52} \\ &= \frac12\int\frac{dx}{\left(x + \frac32\right)^2 + \left(\frac12\right)^2} \end{aligned}

Put $x + \frac32 = y$ ⁠. Then $dx = dy$ ⁠.

Using $\displaystyle\int\frac{dy}{y^2 + a^2} = \frac1a\tan^{-1}\frac ya + C$ ⁠,

\begin{aligned} I_2 &= \frac12\int\frac{dy}{y^2 + \left(\frac12\right)^2} \\ &= \frac12\cdot\frac{1}{\frac12}\tan^{-1}\frac{y}{\frac12} \\ &= \tan^{-1}(2x + 3) \end{aligned}

$\therefore\ I = \frac14\log|2x^2 + 6x + 5|$ $+ \frac12\tan^{-1}(2x + 3) + C$ ⁠.

(ii) Let $I = \displaystyle\int\frac{x + 3}{\sqrt{5 - 4x - x^2}}\,dx$ ⁠.

Let $x + 3 = A(-4 - 2x) + B$ ⁠, where $-4 - 2x = \frac{d}{dx}(5 - 4x - x^2)$ ⁠.

Comparing coefficients: $-2A = 1$ and $-4A + B = 3$ ⁠. So $A = -\frac12$ ⁠, $B = 1$ ⁠.

\begin{aligned} \therefore\ I &= -\frac12\int\frac{(-4 - 2x)\,dx}{\sqrt{5 - 4x - x^2}} \\ &\quad + \int\frac{dx}{\sqrt{5 - 4x - x^2}} \\ &= -\frac12 I_1 + I_2 \end{aligned}

In $I_1$ ⁠, put $5 - 4x - x^2 = t$ ⁠. Then $(-4 - 2x)\,dx = dt$ ⁠.

\begin{aligned} I_1 &= \int\frac{dt}{\sqrt t} = 2\sqrt t \\ &= 2\sqrt{5 - 4x - x^2} \end{aligned}

For $I_2$ ⁠, complete the square:

\begin{aligned} &5 - 4x - x^2 \\ &= -\left[(x + 2)^2 - 4 - 5\right] \\ &= 3^2 - (x + 2)^2 \end{aligned}

Put $x + 2 = y$ ⁠. Then $dx = dy$ ⁠.

Using $\displaystyle\int\frac{dy}{\sqrt{a^2 - y^2}} = \sin^{-1}\frac ya + C$ ⁠,

\begin{aligned} I_2 &= \int\frac{dy}{\sqrt{3^2 - y^2}} = \sin^{-1}\frac y3 \\ &= \sin^{-1}\frac{x + 2}{3} \end{aligned}

$\therefore\ I = -\sqrt{5 - 4x - x^2}$ $+ \sin^{-1}\dfrac{x + 2}{3} + C$ ⁠.

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

The one idea this problem is built on

When the numerator is linear and the denominator is a quadratic $Q$ (or $\sqrt Q$ ⁠), split the numerator as $A\cdot Q' + B$ : a multiple of the derivative of the quadratic, plus a constant (Section 7.4, types (9) and (10)). The $Q'$ part is a perfect substitution ( $t = Q$ ⁠); the constant part $B$ leaves $\int\frac{dx}{Q}$ or $\int\frac{dx}{\sqrt Q}$ ⁠, which Example 9 showed how to do by completing the square.

Given, to find, and the plan

Given: (i) $\frac{x + 2}{2x^2 + 6x + 5}$ ⁠; (ii) $\frac{x + 3}{\sqrt{5 - 4x - x^2}}$ ⁠.
To find: Both indefinite integrals, with where they are valid.
Plan: (1) Find $A$ ⁠, $B$ with numerator $= A\cdot Q' + B$ ⁠. (2) $\int\frac{Q'}{Q}dx = \log|Q|$ or $\int\frac{Q'}{\sqrt Q}dx = 2\sqrt Q$ ⁠. (3) Complete the square for the $B$ part. (4) Combine and differentiate back.

Basics you need

1Comparing coefficients

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

2Substitution $t = Q(x)$

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

3Completing the square

Turns $\int\frac{dx}{Q}$ and $\int\frac{dx}{\sqrt Q}$ into the shapes of Section 7.4 (Example 9). Basics: completing the square

4Formulae 7.4 (3) and (5)

$\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa + C$ ⁠; $\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + C$ ⁠. Lesson: integrals of particular functions

Part (i): $\int\frac{x + 2}{2x^2 + 6x + 5}\,dx$

The domain

The discriminant of $2x^2 + 6x + 5$ is $6^2 - 4\cdot 2\cdot 5 = -4 < 0$ ⁠, and the leading coefficient is positive, so $2x^2 + 6x + 5 > 0$ for every real $x$ ⁠. The integrand is defined everywhere.

A quadratic with negative discriminant has no real roots, so it never changes sign. This also means the modulus in $\log|2x^2 + 6x + 5|$ below changes nothing.
Split the numerator as $A\cdot Q' + B$

$Q = 2x^2 + 6x + 5$ has $Q' = 4x + 6$ ⁠. Write
$x + 2 = A(4x + 6) + B.$
Coefficients of $x$ ⁠: $4A = 1$ ⁠. Constants: $6A + B = 2$ ⁠. So $A = \frac14$ ⁠, $B = 2 - \frac64 = \frac12$ ⁠.

This is formula type 7.4 (9). The identity must hold for every $x$ ⁠, so both the $x$ -coefficients and the constants must agree. Check: $\frac14(4x + 6) + \frac12$ $= x + \frac32 + \frac12 = x + 2$ ⁠.

Type (9), Section 7.4. Basics: splitting fractions
Split the integral into $I_1$ and $I_2$
$\begin{aligned} \int\frac{x + 2}{2x^2 + 6x + 5}\,dx &= \frac14\int\frac{4x + 6}{2x^2 + 6x + 5}\,dx \\ &\quad + \frac12\int\frac{dx}{2x^2 + 6x + 5} \\ &= \frac14 I_1 + \frac12 I_2. \end{aligned}$
Dividing $A(4x + 6) + B$ by the denominator term by term gives two fractions; Property V then splits the integral.

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

Put $t = 2x^2 + 6x + 5$ ⁠, $dt = (4x + 6)\,dx$ ⁠:
$\begin{aligned} I_1 = \int\frac{dt}{t} &= \log|t| + C_1 \\ &= \log|2x^2 + 6x + 5| + C_1. \end{aligned}$
The numerator $4x + 6$ is exactly the derivative of the denominator; that is why we arranged it.

Substitution, Section 7.3.1; formula (xii), Section 7.2. Basics: substitution
$I_2$ by completing the square
$\begin{aligned} I_2 &= \frac12\int\frac{dx}{x^2 + 3x + \frac52} \\ &= \frac12\int\frac{dx}{\left(x + \frac32\right)^2 + \left(\frac12\right)^2}. \end{aligned}$
Put $t = x + \frac32$ ⁠, $dt = dx$ ⁠, and use 7.4 (3) with $a = \frac12$ ⁠:
$\begin{aligned} I_2 &= \frac12\cdot\frac{1}{\frac12}\tan^{-1}\frac{t}{\frac12} + C_2 \\ &= \tan^{-1}2t + C_2 = \tan^{-1}(2x + 3) + C_2. \end{aligned}$
Take out the 2 first. Half of 3 is $\frac32$ ⁠, and $\frac52 - \frac94 = \frac14 = \left(\frac12\right)^2$ ⁠, a positive leftover, so the inverse tangent formula applies. Then $2t = 2x + 3$ ⁠.

Formula 7.4 (3). Basics: completing the square
Combine
$\begin{aligned} \int\frac{x + 2}{2x^2 + 6x + 5}\,dx &= \frac14\log|2x^2 + 6x + 5| \\ &\quad + \frac12\tan^{-1}(2x + 3) + C, \end{aligned}$
with $C = \frac{C_1}{4} + \frac{C_2}{2}$ ⁠.

Part (i), Steps 3, 4 and 5 put together; the constants merge into one.
Check: differentiate the answer

First, $1 + (2x + 3)^2$ $= 4x^2 + 12x + 10$ $= 2(2x^2 + 6x + 5)$ ⁠. Then
$\begin{aligned} &\frac{d}{dx}\left[\frac14\log|Q| + \frac12\tan^{-1}(2x + 3)\right] \\ &= \frac{4x + 6}{4Q} + \frac12\cdot\frac{2}{2Q} \\ &= \frac{x + \frac32 + \frac12}{Q} = \frac{x + 2}{2x^2 + 6x + 5}. \ \checkmark \end{aligned}$
Here $Q = 2x^2 + 6x + 5$ ⁠. The derivative of $\tan^{-1}(2x + 3)$ is $\frac{2}{1 + (2x + 3)^2} = \frac{2}{2Q}$ ⁠, using the identity just shown. Valid for every real $x$ ⁠.

Part (ii): $\int\frac{x + 3}{\sqrt{5 - 4x - x^2}}\,dx$

The domain

$5 - 4x - x^2 = (5 + x)(1 - x)$ ⁠, which is positive exactly when $-5 < x < 1$ ⁠.

The quadratic is under a square root in the denominator, so it must be strictly positive. The product $(5 + x)(1 - x)$ is positive between its roots $-5$ and 1.
Split the numerator as $A\cdot Q' + B$

$Q = 5 - 4x - x^2$ has $Q' = -4 - 2x$ ⁠. Write
$x + 3 = A(-4 - 2x) + B.$
Coefficients of $x$ ⁠: $-2A = 1$ ⁠. Constants: $-4A + B = 3$ ⁠. So $A = -\frac12$ ⁠, $B = 3 + 4A = 1$ ⁠.

This is type 7.4 (10). Check: $-\frac12(-4 - 2x) + 1$ $= 2 + x + 1 = x + 3$ ⁠.

Type (10), Section 7.4. Basics: splitting fractions
Split the integral
$\begin{aligned} \int\frac{x + 3}{\sqrt Q}\,dx &= -\frac12\int\frac{(-4 - 2x)\,dx}{\sqrt{5 - 4x - x^2}} \\ &\quad + \int\frac{dx}{\sqrt{5 - 4x - x^2}} \\ &= -\frac12 I_1 + I_2. \end{aligned}$
The same idea as part (i): one integral built for substitution, one for completing the square.
$I_1$ by substitution

Put $t = 5 - 4x - x^2$ (so $t > 0$ ⁠), $dt = (-4 - 2x)\,dx$ ⁠:
$\begin{aligned} I_1 = \int\frac{dt}{\sqrt t} &= \int t^{-\frac12}\,dt = 2\sqrt t + C_1 \\ &= 2\sqrt{5 - 4x - x^2} + C_1. \end{aligned}$
Power formula with $n = -\frac12$ ⁠: $\frac{t^{1/2}}{1/2} = 2\sqrt t$ ⁠.

Standard formula (i), Section 7.2. Basics: the power rule
$I_2$ by completing the square
$\begin{aligned} 5 - 4x - x^2 &= -\left[(x + 2)^2 - 4 - 5\right] \\ &= 9 - (x + 2)^2 = 3^2 - (x + 2)^2. \end{aligned}$
Put $t = x + 2$ ⁠, $dt = dx$ ⁠, with $-3 < t < 3$ ⁠, and use 7.4 (5):
$\begin{aligned} I_2 = \int\frac{dt}{\sqrt{3^2 - t^2}} &= \sin^{-1}\frac t3 + C_2 \\ &= \sin^{-1}\frac{x + 2}{3} + C_2. \end{aligned}$
$x^2 + 4x = (x + 2)^2 - 4$ ⁠. "Constant minus square" under a root is the $\sin^{-1}$ shape with $a = 3$ ⁠. On the domain $-5 < x < 1$ ⁠, $t = x + 2$ runs over $(-3, 3)$ ⁠, exactly where the formula holds.

Formula 7.4 (5). Basics: completing the square
Combine
$\begin{aligned} \int\frac{x + 3}{\sqrt{5 - 4x - x^2}}\,dx &= -\sqrt{5 - 4x - x^2} \\ &\quad + \sin^{-1}\frac{x + 2}{3} + C, \end{aligned}$
with $C = C_2 - \frac{C_1}{2}$ ⁠.

$-\frac12\cdot 2\sqrt Q = -\sqrt Q$ ⁠. The constants merge, as in the book.
Check: differentiate the answer
$\begin{aligned} \frac{d}{dx}\left(-\sqrt Q\right) &= -\frac{-4 - 2x}{2\sqrt Q} = \frac{x + 2}{\sqrt Q}, \\ \frac{d}{dx}\sin^{-1}\frac{x + 2}{3} &= \frac{\frac13}{\sqrt{1 - \frac{(x + 2)^2}{9}}} = \frac{1}{\sqrt Q}. \end{aligned}$
The sum is $\frac{x + 3}{\sqrt{5 - 4x - x^2}}$ ⁠. $\checkmark$

Here $Q = 5 - 4x - x^2 = 9 - (x + 2)^2$ ⁠. In the second line, multiplying top and bottom by 3 gives $\frac{1}{\sqrt{9 - (x + 2)^2}}$ ⁠. Valid for $-5 < x < 1$ ⁠.

Result

(i) $\displaystyle\int\frac{x + 2}{2x^2 + 6x + 5}\,dx$ $= \frac14\log|2x^2 + 6x + 5|$ $+ \frac12\tan^{-1}(2x + 3) + C$ ⁠, for all real $x$ ⁠.

(ii) $\displaystyle\int\frac{x + 3}{\sqrt{5 - 4x - x^2}}\,dx$ $= -\sqrt{5 - 4x - x^2} + \sin^{-1}\frac{x + 2}{3} + C$ ⁠, for $-5 < x < 1$ ⁠.

These agree with the textbook's solution.

Check it with numbers

By Section 7.8, $\int_a^b f(x)\,dx = F(b) - F(a)$ ⁠. The middle column is a computer's sum of very many thin strips, found without our answers:

Part	Interval	Sum of thin strips	$F(b) - F(a)$
(i)	$[0, 1]$	$0.3010554$	$0.3010554$
(ii)	$[-2, 0]$	$1.4936597$	$3 - \sqrt5 + \sin^{-1}\frac23$ $\approx 1.4936597$

In (i), $F(1) - F(0)$ $= \frac14\log\frac{13}{5}$ $+ \frac12\left(\tan^{-1}5 - \tan^{-1}3\right)$ ⁠.

Common mistakes

Splitting as $x + 2 = A\cdot Q + B$ instead of $A\cdot Q' + B$ ⁠. It must be the derivative $Q'$ ⁠, so that the substitution $t = Q$ works.
Losing the $\frac12$ when taking 2 out of $2x^2 + 6x + 5$ in $I_2$ ⁠.
$\int\frac{dt}{\sqrt t} = \log|t|$ : no, it is $2\sqrt t$ ⁠; the log is for $\int\frac{dt}{t}$ ⁠.
Sign of $A$ in part (ii). $Q' = -4 - 2x$ has a negative $x$ -coefficient, so $A = -\frac12$ ⁠.

What to take away

Linear over quadratic (or over the square root of one): numerator $= A\cdot Q' + B$ ⁠.
The $Q'$ part gives $\log|Q|$ (or $2\sqrt Q$ ⁠); the $B$ part is a completing-the-square integral.
Always check $A$ and $B$ by expanding before integrating.

Every fact used, and where it comes from

Fact	Used in	Source
$px + q = A\frac{d}{dx}(ax^2 + bx + c) + B$	Part (i), Step 2; Part (ii), Step 2	Types (9) and (10), Section 7.4 (Lesson: integrals of particular functions)
$\int\frac{dt}{t} = \log\|t\| + C$ ⁠; $\int t^{-1/2}dt = 2\sqrt t + C$	Part (i), Step 4; Part (ii), Step 4	Standard formulae (xii), (i), Section 7.2
$\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa + C$	Part (i), Step 5	Formula 7.4 (3)
$\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + C$	Part (ii), Step 5	Formula 7.4 (5)
Completing the square; discriminant	Part (i), Steps 1, 5; Part (ii), Step 5	Algebra (Basics: completing the square)
Substitution	Part (i), Step 4; Part (ii), Step 4	Section 7.3.1 (Basics: substitution)

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Example 10: ∫x+22x2+6x+5 dx\int\frac{x + 2}{2x^2 + 6x + 5}\,dx∫2x2+6x+5x+2​dx and ∫x+35−4x−x2 dx\int\frac{x + 3}{\sqrt{5 - 4x - x^2}}\,dx∫5−4x−x2​x+3​dx

Basics you need

Part (i): ∫x+22x2+6x+5 dx\int\frac{x + 2}{2x^2 + 6x + 5}\,dx∫2x2+6x+5x+2​dx

The domain

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 completing the square

Combine

Check: differentiate the answer

Part (ii): ∫x+35−4x−x2 dx\int\frac{x + 3}{\sqrt{5 - 4x - x^2}}\,dx∫5−4x−x2​x+3​dx