Algebra before integrating: division and completing the square

About this lesson

Calculus BC (AP-aligned) › Techniques: substitution

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Algebra before integrating: division and completing the square

Warm-up

Three quick questions from earlier lessons.

Discover

$\dfrac{1}{x^2 + 4x + 5}$ has no obvious inside function. Which rewriting helps?

$\dfrac{1}{(x + 2)^2 + 1}$
$\dfrac{1}{x^2} + \dfrac{1}{4x} + \dfrac{1}{5}$

Key idea

Rewrite first, then integrate.

Divide when the top has a degree at least that of the bottom, or the bottom is one term:

\dfrac{x^2 + 1}{x} = x + \dfrac{1}{x}

. Complete the square to reach

\displaystyle\int \dfrac{du}{u^2 + a^2} = \dfrac{1}{a}\arctan\dfrac{u}{a} + C

.

Worked example

$\displaystyle\int \dfrac{dx}{x^2 + 6x + 13}$ .

x^2 + 6x + 13 = (x + 3)^2 + 4

x² + 6x + 9 + 4.

\displaystyle\int \dfrac{du}{u^2 + 4}

u = x + 3, a = 2.

\tfrac{1}{2}\arctan\dfrac{x + 3}{2} + C

The arctan formula.

Your turn, guided

Your turn: $\displaystyle\int \dfrac{x^2 + 2}{x}\,dx$ for $x > 0$ .

\dfrac{x^2}{2} + 2\ln x + C

x + 2/x.

Practice ★

Find $F$ with $F'(x) = \dfrac{1}{x^2 + 9}$ and $F(0) = 0$ .
Find $F$ with $F'(x) = \dfrac{1}{x^2 + 4}$ and $F(0) = 0$ .

Evaluate $\displaystyle\int_1^2 \dfrac{x^2 + 5x + 1}{x}\,dx$ .
Evaluate $\displaystyle\int_1^2 \dfrac{x^2 + 4x + 1}{x}\,dx$ .

Watch out!

Practice ★★

Find $F$ with $F'(x) = \dfrac{1}{x^2 + 6x + 10}$ and $F(-3) = 0$ .
Find $F$ with $F'(x) = \dfrac{1}{x^2 + 8x + 17}$ and $F(-4) = 0$ .

Exit ticket

Three questions to check you have got it.

Find $F$ with $F'(x) = \dfrac{1}{x^2 + 9}$ and $F(0) = 0$ .
Find $F$ with $F'(x) = \dfrac{1}{x^2 + 4}$ and $F(0) = 0$ .
Find $F$ with $F'(x) = \dfrac{1}{x^2 + 36}$ and $F(0) = 0$ .

Summary card

Algebra first

Divide term by term when the bottom is one term.
Complete the square for $\dfrac{1}{x^2 + bx + c}$ , then arctan.

Watch out: $\dfrac{1}{a + b} \ne \dfrac{1}{a} + \dfrac{1}{b}$ .

End of the lesson

Back to the chapter

Answers: Algebra before integrating: division and completing the square

A

$\frac{\tan^{-1}{\left(\frac{x}{3} \right)}}{3}$
$\frac{\tan^{-1}{\left(\frac{x}{2} \right)}}{2}$

B

$\ln{\left(2 \right)} + \frac{13}{2}$
$\ln{\left(2 \right)} + \frac{11}{2}$

C

$\tan^{-1}{\left(x + 3 \right)}$
$\tan^{-1}{\left(x + 4 \right)}$

D

$\frac{\tan^{-1}{\left(\frac{x}{3} \right)}}{3}$
$\frac{\tan^{-1}{\left(\frac{x}{2} \right)}}{2}$
$\frac{\tan^{-1}{\left(\frac{x}{6} \right)}}{6}$

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