Chapter 7 · Integrals

Exercise 7.4 · Q23: $\int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}\,dx$

Exercise 7.4Integrals of some particular functions · Section 7.4

The question

Integrate the function

\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}.

Exam answer

Let $I = \displaystyle\int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}\,dx$ ⁠.

Let $5x + 3 = A(2x + 4) + B$ ⁠, where $2x + 4 = \frac{d}{dx}(x^2 + 4x + 10)$ ⁠.

Comparing coefficients: $2A = 5$ and $4A + B = 3$ ⁠. So $A = \frac52$ ⁠, $B = -7$ ⁠.

\begin{aligned} \therefore\ I &= \frac52\int\frac{(2x + 4)\,dx}{\sqrt{x^2 + 4x + 10}} \\ &\quad - 7\int\frac{dx}{\sqrt{x^2 + 4x + 10}} \\ &= \frac52 I_1 - 7I_2 \end{aligned}

In $I_1$ ⁠, put $x^2 + 4x + 10 = t$ ⁠. Then $(2x + 4)\,dx = dt$ ⁠.

\begin{aligned} I_1 &= \int\frac{dt}{\sqrt t} = 2\sqrt t \\ &= 2\sqrt{x^2 + 4x + 10} \end{aligned}

For $I_2$ ⁠: $x^2 + 4x + 10 = (x + 2)^2 + \left(\sqrt6\right)^2$ ⁠.

[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)^2 + \left(\sqrt6\right)^2}} \\ &= \log\left|x + 2 + \sqrt{x^2 + 4x + 10}\right| \end{aligned}

$\therefore\ I = 5\sqrt{x^2 + 4x + 10}$ $- 7\log\left|x + 2 + \sqrt{x^2 + 4x + 10}\right|$ $+ C$ ⁠.

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

The one idea this problem is built on

Type 7.4 (10): split the numerator as $A\cdot Q' + B$ with $Q = x^2 + 4x + 10$ ⁠, $Q' = 2x + 4$ ⁠. This gives $5x + 3 = \frac52(2x + 4) - 7$ ⁠. The $Q'$ part becomes $5\sqrt Q$ by substitution. For the constant part, completing the square gives $Q = (x + 2)^2 + \left(\sqrt6\right)^2$ ⁠, "square plus square" under a root, so formula 7.4 (6) gives a log.

Given, to find, and the plan

Given: The integrand $\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}$ ⁠.
To find: $\int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}\,dx$ ⁠, and where the answer is valid.
Plan: (1) Find the domain. (2) Find $A$ ⁠, $B$ with $5x + 3 = A(2x + 4) + B$ ⁠. (3) Split into $I_1$ and $I_2$ ⁠. (4) $I_1$ by substitution. (5) $I_2$ by completing the square and 7.4 (6). (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

3Completing the square

$x^2 + 4x + 10 = (x + 2)^2 + 6$ ⁠. Basics: completing the square

4Formula 7.4 (6)

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

Solution

The domain

$x^2 + 4x + 10 = (x + 2)^2 + 6 \ge 6 > 0$ for every real $x$ ⁠, so the integrand is defined everywhere.

A square is never negative, so adding 6 keeps the quadratic at least 6. (Equivalently, its discriminant $16 - 40 = -24$ is negative.)
Split the numerator as $A\cdot Q' + B$

$Q = x^2 + 4x + 10$ has $Q' = 2x + 4$ ⁠. Write
$5x + 3 = A(2x + 4) + B.$
Coefficients of $x$ ⁠: $2A = 5$ ⁠. Constants: $4A + B = 3$ ⁠. So $A = \frac52$ ⁠, $B = 3 - 10 = -7$ ⁠.

This is type 7.4 (10). Check: $\frac52(2x + 4) - 7 = 5x + 10 - 7 = 5x + 3$ ⁠.

Type 7.4 (10). Basics: splitting fractions
Split the integral into $I_1$ and $I_2$
$\begin{aligned} \int\frac{5x + 3}{\sqrt Q}\,dx &= \frac52\int\frac{(2x + 4)\,dx}{\sqrt{x^2 + 4x + 10}} \\ &\quad - 7\int\frac{dx}{\sqrt{x^2 + 4x + 10}} \\ &= \frac52 I_1 - 7I_2. \end{aligned}$
Dividing $\frac52(2x + 4) - 7$ by $\sqrt Q$ term by term gives two fractions; Property (V) of Section 7.2.1 splits the integral.

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

Put $t = x^2 + 4x + 10$ (so $t > 0$ ⁠), $dt = (2x + 4)\,dx$ ⁠:
$\begin{aligned} I_1 = \int\frac{dt}{\sqrt t} &= 2\sqrt t + C_1 \\ &= 2\sqrt{x^2 + 4x + 10} + C_1. \end{aligned}$
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 completing the square

$x^2 + 4x + 10 = (x + 2)^2 + \left(\sqrt6\right)^2$ ⁠. Put $t = x + 2$ ⁠, $dt = dx$ ⁠, and use 7.4 (6) with $a = \sqrt6$ ⁠:
$\begin{aligned} I_2 &= \int\frac{dt}{\sqrt{t^2 + \left(\sqrt6\right)^2}} = \log\left|t + \sqrt{t^2 + 6}\right| + C_2 \\ &= \log\left|x + 2 + \sqrt{x^2 + 4x + 10}\right| + C_2. \end{aligned}$
Half of 4 is 2, and $10 - 4 = 6$ ⁠, a positive leftover, so this is "square plus square": formula 7.4 (6). Here $t + \sqrt{t^2 + 6} > 0$ for every $t$ (the root is bigger than $|t|$ ⁠), so the modulus changes nothing.

Type 7.4 (8); formula 7.4 (6). Basics: completing the square
Combine
$\begin{aligned} \int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}\,dx &= 5\sqrt{x^2 + 4x + 10} \\ &\quad - 7\log\left|x + 2 + \sqrt{x^2 + 4x + 10}\right| \\ &\quad + C, \end{aligned}$
with $C = \frac52 C_1 - 7C_2$ ⁠.

$\frac52\cdot 2\sqrt Q = 5\sqrt Q$ ⁠. The constants merge into one. This is exactly the textbook's form.
Check: differentiate the answer

Write $s = \sqrt{x^2 + 4x + 10}$ ⁠, so $s' = \frac{2x + 4}{2s} = \frac{x + 2}{s}$ ⁠. Then
$\begin{aligned} \frac{d}{dx}\log|x + 2 + s| &= \frac{1 + \frac{x + 2}{s}}{x + 2 + s} = \frac1s, \\ \frac{d}{dx}\left(5s\right) &= \frac{5(x + 2)}{s}. \end{aligned}$
So the derivative of our answer is $\frac{5(x + 2)}{s} - \frac7s = \frac{5x + 3}{s}$ ⁠. $\checkmark$

In the first line, $1 + \frac{x + 2}{s} = \frac{s + x + 2}{s}$ ⁠, and the factor $x + 2 + s$ (never 0) cancels. The result is the integrand, for every real $x$ ⁠.

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

Result

\begin{aligned} \int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}\,dx &= 5\sqrt{x^2 + 4x + 10} \\ &\quad - 7\log\left|x + 2 + \sqrt{x^2 + 4x + 10}\right| \\ &\quad + C, \end{aligned}

valid for every real $x$ ⁠. (Textbook answer: $5\sqrt{x^2 + 4x + 10}$ $- 7\log\left|x + 2 + \sqrt{x^2 + 4x + 10}\right| + C$ ⁠, the same.)

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:

Interval	Sum of thin strips	$F(b) - F(a)$
$[0, 1]$	$1.5499858$	$1.5499858$
$[-3, -1]$	$-5.5675582$	$-5.5675582$

On $[0, 1]$ ⁠, $F(1) - F(0) = 5\left(\sqrt{15} - \sqrt{10}\right)$ $- 7\log\frac{3 + \sqrt{15}}{2 + \sqrt{10}}$ ⁠.

Common mistakes

Wrong $B$ ⁠. $B = 3 - 4A = 3 - 10 = -7$ ⁠; forgetting the $4A$ gives $B = 3$ ⁠.
Forgetting that $\frac52\cdot 2 = 5$ ⁠. The $\sqrt Q$ term has coefficient 5, not $\frac52$ ⁠.
Writing $\log\left|x + \sqrt{x^2 + 4x + 10}\right|$ ⁠. Undo the shift $t = x + 2$ fully: the log contains $x + 2$ ⁠.

What to take away

Numerator $= A\cdot Q' + B$ ⁠; the $Q'$ part gives $2A\sqrt Q$ ⁠, the $B$ part a completing-the-square integral.
A positive leftover under a root (with $+x^2$ ⁠) means formula 7.4 (6), valid for every $x$ ⁠.

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	Steps 4, 5	Section 7.3.1 (Basics: substitution)
Completing the square	Steps 1, 5	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 5	Formula 7.4 (6)
$\frac{d}{dx}\log\|x\| = \frac1x$ ⁠; 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 · Q23: ∫5x+3x2+4x+10 dx\int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}\,dx∫x2+4x+10​5x+3​dx

Basics you need

Solution

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

Every fact used, and where it comes from

Exercise 7.4 · Q23: $\int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}\,dx$

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

Split the integral into $I_1$ and $I_2$

$I_1$ by substitution

$I_2$ by completing the square