Chapter 7 · Integrals

Exercise 7.4 · Q16: $\int\frac{4x + 1}{\sqrt{2x^2 + x - 3}}\,dx$

Exercise 7.4Integrals of some particular functions · Section 7.4

The question

Integrate the function

\frac{4x + 1}{\sqrt{2x^2 + x - 3}}.

Exam answer

Let $I = \displaystyle\int\frac{4x + 1}{\sqrt{2x^2 + x - 3}}\,dx$ ⁠.

Put $2x^2 + x - 3 = t$ ⁠. Then $(4x + 1)\,dx = dt$ ⁠.

I = \int\frac{dt}{\sqrt t} = 2\sqrt t + C

$\therefore\ I = 2\sqrt{2x^2 + x - 3} + C$ ⁠.

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

The one idea this problem is built on

This is type 7.4 (10), $\int\frac{px + q}{\sqrt{ax^2 + bx + c}}\,dx$ ⁠, where we write the numerator as $A\cdot Q' + B$ ⁠. Here the numerator is exactly the derivative of the quadratic: $\frac{d}{dx}(2x^2 + x - 3) = 4x + 1$ ⁠, so $A = 1$ ⁠, $B = 0$ ⁠. The substitution $t = 2x^2 + x - 3$ then turns the integral into $\int t^{-1/2}\,dt$ ⁠.

Given, to find, and the plan

Given: The integrand $\frac{4x + 1}{\sqrt{2x^2 + x - 3}}$ ⁠.
To find: $\int\frac{4x + 1}{\sqrt{2x^2 + x - 3}}\,dx$ ⁠, and where the answer is valid.
Plan: (1) Find the domain. (2) Notice that the numerator is the derivative of $Q = 2x^2 + x - 3$ ⁠. (3) Substitute $t = Q$ and use the power formula. (4) Differentiate back.

Basics you need

1Factorising a quadratic

$2x^2 + x - 3 = (2x + 3)(x - 1)$ ⁠, used to find where it is positive. Basics: completing the square

2Substitution $t = Q(x)$

If $t = Q(x)$ then $dt = Q'(x)\,dx$ ⁠, so $\int\frac{Q'(x)}{\sqrt{Q(x)}}\,dx = \int\frac{dt}{\sqrt t}$ ⁠. Basics: substitution

3The power formula

$\int t^n\,dt = \frac{t^{n + 1}}{n + 1} + C$ for $n \ne -1$ ⁠; with $n = -\frac12$ it gives $2\sqrt t$ ⁠. Basics: the power rule

Solution

Where the integrand exists

$2x^2 + x - 3 = (2x + 3)(x - 1)$ ⁠, which is positive when both factors have the same sign:
$x < -\frac32 \quad\text{or}\quad x > 1.$
The quadratic is under a square root in a denominator, so it must be strictly positive. The roots are $-\frac32$ and 1; an upward parabola is positive outside its roots. Check the factorisation: $(2x + 3)(x - 1)$ $= 2x^2 - 2x + 3x - 3$ $= 2x^2 + x - 3$ ⁠.
Compare the numerator with $Q'$

Let $Q = 2x^2 + x - 3$ ⁠. Then $Q' = 4x + 1$ ⁠, which is the numerator. In the language of type 7.4 (10),
$4x + 1 = A(4x + 1) + B \ \text{ with } A = 1,\ B = 0.$
The book's method for $\int\frac{px + q}{\sqrt{ax^2 + bx + c}}\,dx$ is to write the numerator as a multiple of the derivative of the quadratic plus a constant. Always try this first: here there is no constant left over, so only the substitution part is needed and no completing the square.

Type 7.4 (10). Basics: splitting fractions
Substitute $t = 2x^2 + x - 3$

Put $t = 2x^2 + x - 3$ (so $t > 0$ on the domain); then $dt = (4x + 1)\,dx$ and
$\begin{aligned} \int\frac{4x + 1}{\sqrt{2x^2 + x - 3}}\,dx &= \int\frac{dt}{\sqrt t} = \int t^{-\frac12}\,dt \\ &= \frac{t^{\frac12}}{\frac12} + C = 2\sqrt t + C \\ &= 2\sqrt{2x^2 + x - 3} + C. \end{aligned}$
The whole of $(4x + 1)\,dx$ becomes $dt$ ⁠, which is why we checked that the numerator is $Q'$ ⁠. The power formula with $n = -\frac12$ gives $n + 1 = \frac12$ ⁠, and dividing by $\frac12$ is multiplying by 2.

Substitution, Section 7.3.1; standard formula (i), Section 7.2. Basics: the power rule
Check: differentiate the answer

By the chain rule, with inner function $Q = 2x^2 + x - 3$ ⁠:
$\begin{aligned} \frac{d}{dx}\left(2\sqrt Q\right) &= 2\cdot\frac{Q'}{2\sqrt Q} \\ &= \frac{4x + 1}{\sqrt{2x^2 + x - 3}}. \ \checkmark \end{aligned}$
$\frac{d}{dQ}\sqrt Q = \frac{1}{2\sqrt Q}$ ⁠, times the inner derivative $Q' = 4x + 1$ ⁠. Valid wherever $Q > 0$ ⁠, on both pieces of the domain.

Chain rule (Theorem 4, Section 5.3.1). Basics: integration undoes differentiation

Result

\int\frac{4x + 1}{\sqrt{2x^2 + x - 3}}\,dx = 2\sqrt{2x^2 + x - 3} + C,

valid for $x < -\frac32$ or $x > 1$ ⁠. (Textbook answer: $2\sqrt{2x^2 + x - 3} + C$ ⁠.)

Check it with numbers

By Section 7.8, $\int_a^b f(x)\,dx = F(b) - F(a)$ with $F(x) = 2\sqrt{2x^2 + x - 3}$ ⁠. 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	$F(b) - F(a)$
$[2, 3]$	$3.1937788$	$6\sqrt2 - 2\sqrt7 = 3.1937788$
$[-3, -2]$	$-3.4641016$	$2\sqrt3 - 4\sqrt3 = -3.4641016$

The second value is negative because the numerator $4x + 1$ is negative there.

Common mistakes

Completing the square without looking at the numerator. It works, but it is much longer; check first whether the numerator is (a multiple of) $Q'$ ⁠.
Writing $\int\frac{dt}{\sqrt t} = \log|t|$ ⁠. The log is for $\int\frac{dt}{t}$ ⁠; here $\int t^{-1/2}\,dt = 2\sqrt t$ ⁠.
Forgetting the factor 2. $\frac{t^{1/2}}{1/2} = 2t^{1/2}$ ⁠, not $\frac12 t^{1/2}$ ⁠.

What to take away

$\int\frac{Q'(x)}{\sqrt{Q(x)}}\,dx = 2\sqrt{Q(x)} + C$ ⁠.
For $\int\frac{px + q}{\sqrt Q}\,dx$ ⁠, split the numerator as $A\cdot Q' + B$ ⁠; when $B = 0$ the answer is just $2A\sqrt Q + C$ ⁠.

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)
Substitution $t = 2x^2 + x - 3$ ⁠, $dt = (4x + 1)\,dx$	Step 3	Section 7.3.1 (Basics: substitution)
$\int t^n\,dt = \frac{t^{n + 1}}{n + 1} + C$ ⁠, $n \ne -1$	Step 3	Standard formula (i), Section 7.2
Chain rule	Step 4	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 · Q16: ∫4x+12x2+x−3 dx\int\frac{4x + 1}{\sqrt{2x^2 + x - 3}}\,dx∫2x2+x−3​4x+1​dx

Basics you need

Solution

Where the integrand exists

Compare the numerator with Q′Q'Q′

Substitute t=2x2+x−3t = 2x^2 + x - 3t=2x2+x−3

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.4 · Q16: $\int\frac{4x + 1}{\sqrt{2x^2 + x - 3}}\,dx$

Compare the numerator with $Q'$

Substitute $t = 2x^2 + x - 3$