Chapter 7 · Integrals

Exercise 7.4 · Q18: $\int\frac{5x - 2}{1 + 2x + 3x^2}\,dx$

Exercise 7.4Integrals of some particular functions · Section 7.4

The question

Integrate the function

\frac{5x - 2}{1 + 2x + 3x^2}.

Exam answer

Let $I = \displaystyle\int\frac{5x - 2}{1 + 2x + 3x^2}\,dx$ ⁠.

Let $5x - 2 = A(6x + 2) + B$ ⁠, where $6x + 2 = \frac{d}{dx}(3x^2 + 2x + 1)$ ⁠.

Comparing coefficients: $6A = 5$ and $2A + B = -2$ ⁠. So $A = \frac56$ ⁠, $B = -\frac{11}{3}$ ⁠.

\begin{aligned} \therefore\ I &= \frac56\int\frac{(6x + 2)\,dx}{3x^2 + 2x + 1} \\ &\quad - \frac{11}{3}\int\frac{dx}{3x^2 + 2x + 1} \\ &= \frac56 I_1 - \frac{11}{3}I_2 \end{aligned}

In $I_1$ ⁠, put $3x^2 + 2x + 1 = t$ ⁠. Then $(6x + 2)\,dx = dt$ ⁠.

\begin{aligned} I_1 &= \int\frac{dt}{t} = \log|t| \\ &= \log\left|3x^2 + 2x + 1\right| \end{aligned}

For $I_2$ ⁠, complete the square:

\begin{aligned} &3x^2 + 2x + 1 \\ &= 3\left[\left(x + \frac13\right)^2 + \frac29\right] \\ &= 3\left[\left(x + \frac13\right)^2 + \left(\frac{\sqrt2}{3}\right)^2\right] \end{aligned}

[Using $\displaystyle\int\frac{dx}{x^2 + a^2}$ $= \frac1a\tan^{-1}\frac xa + C$ ]

\begin{aligned} I_2 &= \frac13\int\frac{dx}{\left(x + \frac13\right)^2 + \left(\frac{\sqrt2}{3}\right)^2} \\ &= \frac13\cdot\frac{3}{\sqrt2}\tan^{-1}\frac{x + \frac13}{\frac{\sqrt2}{3}} \\ &= \frac{1}{\sqrt2}\tan^{-1}\frac{3x + 1}{\sqrt2} \end{aligned}

$\therefore\ I = \dfrac56\log\left|3x^2 + 2x + 1\right|$ $- \dfrac{11}{3\sqrt2}\tan^{-1}\dfrac{3x + 1}{\sqrt2}$ $+ C$ ⁠.

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

The one idea this problem is built on

Linear over quadratic is type 7.4 (9): write the numerator as $A\cdot Q' + B$ ⁠, where $Q = 3x^2 + 2x + 1$ and $Q' = 6x + 2$ ⁠. The $Q'$ part integrates to $\log|Q|$ by substitution. The constant part leaves $\int\frac{dx}{Q}$ ⁠, and since $Q$ has no real roots, completing the square gives "square plus a positive constant": an inverse tangent by formula 7.4 (3).

Given, to find, and the plan

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

$\int\frac{Q'(x)}{Q(x)}\,dx = \log|Q(x)| + C$ ⁠. Basics: logarithms and 1/x

3Completing the square

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

4Formula 7.4 (3)

$\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa + C$ for $a > 0$ ⁠. Lesson: integrals of particular functions

Solution

The domain

The denominator is $Q = 3x^2 + 2x + 1$ ⁠. Its discriminant is $2^2 - 4\cdot 3\cdot 1 = -8 < 0$ and its leading coefficient is positive, so $Q > 0$ for every real $x$ ⁠. The integrand is defined everywhere.

A quadratic with negative discriminant has no real roots, so it never changes sign; with a positive $x^2$ coefficient it is always positive. So the modulus in $\log|Q|$ below changes nothing.
Split the numerator as $A\cdot Q' + B$

$Q' = 6x + 2$ ⁠. Write
$5x - 2 = A(6x + 2) + B.$
Coefficients of $x$ ⁠: $6A = 5$ ⁠. Constants: $2A + B = -2$ ⁠. So $A = \frac56$ ⁠, $B = -2 - \frac53 = -\frac{11}{3}$ ⁠.

This is type 7.4 (9). Check: $\frac56(6x + 2) - \frac{11}{3} = 5x + \frac53 - \frac{11}{3} = 5x - 2$ ⁠.

Type 7.4 (9). Basics: splitting fractions
Split the integral into $I_1$ and $I_2$
$\begin{aligned} \int\frac{5x - 2}{3x^2 + 2x + 1}\,dx &= \frac56\int\frac{6x + 2}{3x^2 + 2x + 1}\,dx \\ &\quad - \frac{11}{3}\int\frac{dx}{3x^2 + 2x + 1} \\ &= \frac56 I_1 - \frac{11}{3}I_2. \end{aligned}$
Dividing $A(6x + 2) + B$ by $Q$ term by term gives two fractions; Property (V) of Section 7.2.1 splits the integral. We have also reordered the denominator as $3x^2 + 2x + 1$ ⁠.

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

Put $t = 3x^2 + 2x + 1$ ⁠, $dt = (6x + 2)\,dx$ ⁠:
$I_1 = \int\frac{dt}{t} = \log|t| + C_1 = \log\left|3x^2 + 2x + 1\right| + C_1.$
The numerator $6x + 2$ is exactly the derivative of the denominator; that is why we arranged it in Step 2.

Substitution, Section 7.3.1; standard formula (xii), Section 7.2. Basics: substitution
$I_2$ by completing the square
$\begin{aligned} 3x^2 + 2x + 1 &= 3\left(x^2 + \frac{2x}{3} + \frac13\right) \\ &= 3\left[\left(x + \frac13\right)^2 + \frac29\right]. \end{aligned}$
Let $a = \frac{\sqrt2}{3}$ ⁠, so $a^2 = \frac29$ ⁠. Put $t = x + \frac13$ ⁠, $dt = dx$ ⁠, and use 7.4 (3):
$\begin{aligned} I_2 &= \frac13\int\frac{dt}{t^2 + a^2} = \frac13\cdot\frac1a\tan^{-1}\frac ta + C_2 \\ &= \frac13\cdot\frac{3}{\sqrt2}\tan^{-1}\frac{3x + 1}{\sqrt2} + C_2 \\ &= \frac{1}{\sqrt2}\tan^{-1}\frac{3x + 1}{\sqrt2} + C_2. \end{aligned}$
Take out the 3 first so that $x^2$ has coefficient 1. Half of $\frac23$ is $\frac13$ ⁠, and $\frac13 - \frac19 = \frac29$ ⁠, a positive leftover, so the inverse tangent formula applies. Then $\frac ta = \frac{x + \frac13}{\frac{\sqrt2}{3}} = \frac{3x + 1}{\sqrt2}$ and $\frac1a = \frac{3}{\sqrt2}$ ⁠.

Type 7.4 (7); formula 7.4 (3). Basics: completing the square
Combine
$\begin{aligned} \int\frac{5x - 2}{1 + 2x + 3x^2}\,dx &= \frac56\log\left|3x^2 + 2x + 1\right| \\ &\quad - \frac{11}{3\sqrt2}\tan^{-1}\frac{3x + 1}{\sqrt2} + C, \end{aligned}$
with $C = \frac56 C_1 - \frac{11}{3}C_2$ ⁠.

$\frac{11}{3}\cdot\frac{1}{\sqrt2} = \frac{11}{3\sqrt2}$ ⁠. The constants merge into one. This is exactly the textbook's form.
Check: differentiate the answer

Write $Q = 3x^2 + 2x + 1$ ⁠. First, $2 + (3x + 1)^2 = 9x^2 + 6x + 3 = 3Q$ ⁠. Then
$\begin{aligned} \frac{d}{dx}\tan^{-1}\frac{3x + 1}{\sqrt2} &= \frac{\frac{3}{\sqrt2}}{1 + \frac{(3x + 1)^2}{2}} = \frac{\frac{6}{\sqrt2}}{3Q} = \frac{\sqrt2}{Q}, \\ \frac{d}{dx}\left[\frac56\log|Q|\right] &= \frac{5(6x + 2)}{6Q} = \frac{5x + \frac53}{Q}. \end{aligned}$
So the derivative of our answer is
$\frac{5x + \frac53}{Q} - \frac{11}{3\sqrt2}\cdot\frac{\sqrt2}{Q} = \frac{5x + \frac53 - \frac{11}{3}}{Q} = \frac{5x - 2}{Q}. \ \checkmark$
For the inverse tangent, the chain rule gives the inner derivative $\frac{3}{\sqrt2}$ ⁠; multiplying the top and bottom by 2 gives $\frac{6/\sqrt2}{2 + (3x + 1)^2}$ ⁠, and $\frac{6}{\sqrt2} = 3\sqrt2$ ⁠. Valid for every real $x$ ⁠.

Chain rule (Theorem 4, Section 5.3.1); derivatives (x), (xii), Section 7.2. Basics: integration undoes differentiation

Result

\begin{aligned} \int\frac{5x - 2}{1 + 2x + 3x^2}\,dx &= \frac56\log\left|3x^2 + 2x + 1\right| \\ &\quad - \frac{11}{3\sqrt2}\tan^{-1}\frac{3x + 1}{\sqrt2} + C, \end{aligned}

valid for every real $x$ ⁠. (Textbook answer: $\frac56\log\left|3x^2 + 2x + 1\right|$ $- \frac{11}{3\sqrt2}\tan^{-1}\left(\frac{3x + 1}{\sqrt2}\right) + C$ ⁠, 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	$F(b) - F(a)$
$[0, 1]$	$-0.1026367$	$-0.1026367$
$[-1, 0]$	$-4.6502653$	$-4.6502653$

On $[0, 1]$ ⁠, $F(1) - F(0) = \frac56\log 6$ $- \frac{11}{3\sqrt2}\left(\tan^{-1}2\sqrt2 - \tan^{-1}\frac{1}{\sqrt2}\right)$ ⁠.

Common mistakes

Splitting as $A\cdot Q + B$ instead of $A\cdot Q' + B$ ⁠. It must be the derivative $Q' = 6x + 2$ ⁠.
Wrong $B$ ⁠. $B = -2 - 2A = -2 - \frac53 = -\frac{11}{3}$ ⁠; forgetting the $2A$ gives $B = -2$ ⁠.
Losing the $\frac13$ when taking 3 out of $3x^2 + 2x + 1$ in $I_2$ ⁠.
Using $a = \frac29$ instead of $a = \frac{\sqrt2}{3}$ in 7.4 (3).

What to take away

Linear over quadratic: numerator $= A\cdot Q' + B$ ⁠; the $Q'$ part gives $A\log|Q|$ ⁠.
When $Q$ has negative discriminant, $\int\frac{dx}{Q}$ is an inverse tangent after completing the square.
Check $A$ and $B$ by expanding before integrating.

Every fact used, and where it comes from

Fact	Used in	Source
$px + q = A\frac{d}{dx}(ax^2 + bx + c) + B$	Step 2	Type 7.4 (9) (Lesson: integrals of particular functions)
Integral of a sum of constant multiples	Step 3	Property (V), Section 7.2.1
$\int\frac{dt}{t} = \log\|t\| + C$	Step 4	Standard formula (xii), Section 7.2
Substitution	Steps 4, 5	Section 7.3.1 (Basics: substitution)
Completing the square; discriminant	Steps 1, 5	Type 7.4 (7) (Basics: completing the square)
$\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa + C$	Step 5	Formula 7.4 (3)
$\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$ ⁠; chain rule	Step 7	Derivative (x), 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 · Q18: ∫5x−21+2x+3x2 dx\int\frac{5x - 2}{1 + 2x + 3x^2}\,dx∫1+2x+3x25x−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 · Q18: $\int\frac{5x - 2}{1 + 2x + 3x^2}\,dx$

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