Chapter 7 · Integrals

Exercise 7.4 · Q21: $\int\frac{x + 2}{\sqrt{x^2 + 2x + 3}}\,dx$

Exercise 7.4Integrals of some particular functions · Section 7.4

The question

Integrate the function

\frac{x + 2}{\sqrt{x^2 + 2x + 3}}.

Exam answer

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

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

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

\begin{aligned} \therefore\ I &= \frac12\int\frac{(2x + 2)\,dx}{\sqrt{x^2 + 2x + 3}} \\ &\quad + \int\frac{dx}{\sqrt{x^2 + 2x + 3}} \\ &= \frac12 I_1 + I_2 \end{aligned}

In $I_1$ ⁠, put $x^2 + 2x + 3 = t$ ⁠. Then $(2x + 2)\,dx = dt$ ⁠.

\begin{aligned} I_1 &= \int\frac{dt}{\sqrt t} = 2\sqrt t \\ &= 2\sqrt{x^2 + 2x + 3} \end{aligned}

For $I_2$ ⁠: $x^2 + 2x + 3 = (x + 1)^2 + \left(\sqrt2\right)^2$ ⁠.

[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 + 1)^2 + \left(\sqrt2\right)^2}} \\ &= \log\left|x + 1 + \sqrt{x^2 + 2x + 3}\right| \end{aligned}

$\therefore\ I = \sqrt{x^2 + 2x + 3}$ $+ \log\left|x + 1 + \sqrt{x^2 + 2x + 3}\right|$ $+ C$ ⁠.

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 $Q = x^2 + 2x + 3$ ⁠, $Q' = 2x + 2$ ⁠. This gives $x + 2 = \frac12(2x + 2) + 1$ ⁠. The $Q'$ part becomes $\sqrt Q$ by substitution. For the constant part, completing the square gives $Q = (x + 1)^2 + \left(\sqrt2\right)^2$ ⁠, "square plus constant", so formula 7.4 (6) gives a log.

Given, to find, and the plan

Given: The integrand $\frac{x + 2}{\sqrt{x^2 + 2x + 3}}$ ⁠.
To find: $\int\frac{x + 2}{\sqrt{x^2 + 2x + 3}}\,dx$ ⁠, and where the answer is valid.
Plan: (1) Find the domain. (2) Find $A$ ⁠, $B$ with $x + 2 = A(2x + 2) + B$ ⁠. (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

$x^2 + 2x + 3 = (x + 1)^2 + 2$ ⁠. 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

$x^2 + 2x + 3 = (x + 1)^2 + 2 \ge 2 > 0$ for every real $x$ ⁠, so the integrand is defined everywhere.

A square is never negative, so adding 2 keeps the quadratic at least 2. (Equivalently, its discriminant $4 - 12 = -8$ is negative.)
Split the numerator as $A\cdot Q' + B$

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

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

Type 7.4 (10). Basics: splitting fractions
Split the integral into $I_1$ and $I_2$
$\begin{aligned} \int\frac{x + 2}{\sqrt Q}\,dx &= \frac12\int\frac{(2x + 2)\,dx}{\sqrt{x^2 + 2x + 3}} \\ &\quad + \int\frac{dx}{\sqrt{x^2 + 2x + 3}} \\ &= \frac12 I_1 + I_2. \end{aligned}$
Dividing $\frac12(2x + 2) + 1$ 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 $t = x^2 + 2x + 3$ (so $t > 0$ ⁠), $dt = (2x + 2)\,dx$ ⁠:
$\begin{aligned} I_1 = \int\frac{dt}{\sqrt t} &= 2\sqrt t + C_1 \\ &= 2\sqrt{x^2 + 2x + 3} + C_1. \end{aligned}$
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

$x^2 + 2x + 3 = (x + 1)^2 + \left(\sqrt2\right)^2$ ⁠. Put $t = x + 1$ ⁠, $dt = dx$ ⁠, and use 7.4 (6) with $a = \sqrt2$ ⁠:
$\begin{aligned} I_2 &= \int\frac{dt}{\sqrt{t^2 + \left(\sqrt2\right)^2}} = \log\left|t + \sqrt{t^2 + 2}\right| + C_2 \\ &= \log\left|x + 1 + \sqrt{x^2 + 2x + 3}\right| + C_2. \end{aligned}$
Half of 2 is 1, and $3 - 1 = 2$ ⁠, a positive leftover, so this is "square plus square": formula 7.4 (6). Here $t + \sqrt{t^2 + 2} > 0$ 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
$\begin{aligned} \int\frac{x + 2}{\sqrt{x^2 + 2x + 3}}\,dx &= \sqrt{x^2 + 2x + 3} \\ &\quad + \log\left|x + 1 + \sqrt{x^2 + 2x + 3}\right| \\ &\quad + C, \end{aligned}$
with $C = \frac{C_1}{2} + C_2$ ⁠.

$\frac12\cdot 2\sqrt Q = \sqrt Q$ ⁠. The constants merge into one. This is exactly the textbook's form.
Check: differentiate the answer

Write $s = \sqrt{x^2 + 2x + 3}$ ⁠, so $s' = \frac{2x + 2}{2s} = \frac{x + 1}{s}$ ⁠. Then
$\begin{aligned} \frac{d}{dx}\log|x + 1 + s| &= \frac{1 + \frac{x + 1}{s}}{x + 1 + s} = \frac1s, \\ \frac{d}{dx}\left(s + \log|x + 1 + s|\right) &= \frac{x + 1}{s} + \frac1s = \frac{x + 2}{s}. \ \checkmark \end{aligned}$
In the first line, $1 + \frac{x + 1}{s} = \frac{s + x + 1}{s}$ ⁠, and the factor $x + 1 + s$ (never 0) cancels. The result $\frac{x + 2}{s}$ 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{x + 2}{\sqrt{x^2 + 2x + 3}}\,dx &= \sqrt{x^2 + 2x + 3} \\ &\quad + \log\left|x + 1 + \sqrt{x^2 + 2x + 3}\right| \\ &\quad + C, \end{aligned}

valid for every real $x$ ⁠. (Textbook answer: $\sqrt{x^2 + 2x + 3}$ $+ \log\left|x + 1 + \sqrt{x^2 + 2x + 3}\right| + C$ ⁠, 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	$F(b) - F(a)$
$[0, 1]$	$1.2051758$	$1.2051758$
$[-3, -2]$	$-0.2297020$	$-0.2297020$

On $[0, 1]$ ⁠, $F(1) - F(0) = \sqrt6 - \sqrt3 + \log\frac{2 + \sqrt6}{1 + \sqrt3}$ ⁠.

Common mistakes

Wrong $B$ ⁠. $B = 2 - 2A = 1$ ⁠; forgetting the $2A$ gives $B = 2$ ⁠.
Using $\tan^{-1}$ ⁠. $\tan^{-1}$ belongs to $\int\frac{dx}{x^2 + a^2}$ (no root). With a root, "square plus square" gives the log of 7.4 (6).
Writing $\log\left|x + \sqrt{x^2 + 2x + 3}\right|$ ⁠. The shift $t = x + 1$ must be undone fully: the log contains $x + 1$ ⁠.

What to take away

Numerator $= A\cdot Q' + B$ ⁠; the $Q'$ part gives $2A\sqrt Q$ ⁠.
For every real $x$ ⁠, $\begin{aligned} &\int\frac{dx}{\sqrt{(x + h)^2 + a^2}} \\ &= \log\left|x + h + \sqrt{(x + h)^2 + a^2}\right| + C. \end{aligned}$

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)

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 7.4 · Q21: ∫x+2x2+2x+3 dx\int\frac{x + 2}{\sqrt{x^2 + 2x + 3}}\,dx∫x2+2x+3​x+2​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 · Q21: $\int\frac{x + 2}{\sqrt{x^2 + 2x + 3}}\,dx$

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