Home / AP Calculus AB and BC: Chapter 6 – Techniques of Integration : 6.1 – Basic Integration Rules Study Notes

AP Calculus AB and BC: Chapter 6 – Techniques of Integration : 6.1 – Basic Integration Rules Study Notes

 Basic Integration Rules

In this section, we will study several integration techniques for fitting an integrand into one of the basic integration rules. The basic integration rules are reviewed in Table 6.1 on page 252 .

Procedures for Fitting Integrands to Basic Rules

Procedure                                                                                                                  Example

1. Separating numerator                                                                                  $\frac{1-2 x}{1+x^2}=\frac{1}{1+x^2}-\frac{2 x}{1+x^2}$

2. Adding and subtracting terms in numerator                                      $\frac{1}{1-e^x}=\frac{1-e^x+e^x}{1-e^x}=\frac{1-e^x}{1-e^x}+\frac{e^x}{1-e^x}$

3. Dividing improper fractions                                                                       $\frac{x^3-3 x}{x^2-1}=x-\frac{2 x}{x^2-1}$

4. Completing the square                                                                                $\frac{1}{\sqrt{4 x-x^2}}=\frac{1}{\sqrt{4-(x-2)^2}}$

Other integration techniques, such as the simple substitution method, were covered in section 4.8. Using trigonometric identities, trigonometric substitution, Method of Partial Fractions and Integration by Parts will be covered later in this chapter.

1. Separating numerator

Example 1

  • Evaluate $\int \frac{1-2 x}{1+x^2} d x$.
▶️Answer/Explanation

Solution

$\int \frac{1-2 x}{1+x^2} d x=\int \frac{1}{1+x^2} d x+\int \frac{-2 x}{1+x^2} d x$                                                Separate the numerator

$\int \frac{1}{1+x^2} d x=\arctan x$                                                                                                                                   Basic integration rules

$\int \frac{-2 x}{1+x^2} d x=\int \frac{-d u}{u}=-\ln u$                                                                                                  Let $u=1+x^2$, then $d u=2 x d x$.

Therefore $\int \frac{1-2 x}{1+x^2} d x=\arctan x-\ln \left(1+x^2\right)+C$.

Differentiation Rules and Basic Integration Rules

Adding and subtracting terms in numerator

Example 2

  • Evaluate $\int \frac{1}{1-e^x} d x$.
▶️Answer/Explanation

Solution   
$\int \frac{1}{1-e^x} d x=\int \frac{1-e^x+e^x}{1-e^x} d x$                                          Add and subtract $e^x$ in the numerator.

$=\int \frac{1-e^x}{1-e^x} d x+\int \frac{e^x}{1-e^x} d x$                                                              Separate the numerator

$=\int d x+\int \frac{e^x}{1-e^x} d x$

$=x-\ln \left(1-e^x\right)+C$                                                                                                                        Use the basic integration rules.

Dividing improper fractions

Example 3

  • Evaluate $\int \frac{x^3-3 x}{x^2-1} d x$.
▶️Answer/Explanation

Solution
$
\int \frac{x^3-3 x}{x^2-1} d x=\int\left(x-\frac{2 x}{x^2-1}\right) d x
$                                                                                                                                                                Divide an improper fraction.

$=\int x d x-\int \frac{2 x}{x^2-1} d x$

$=\frac{1}{2} x^2-\ln \left(x^2-1\right)+C $ Use the basic integration rules.

Example4

  • $
    \int \frac{1+\sin x}{\cos ^2 x} d x=
    $

(A) $\tan x-\sec x \tan x+C$

(B) $\tan x+\sec x+C$

(C) $\tan x+\sec ^2 x+C$

(D) $\ln \left(1+\cos ^2 x\right)+C$

▶️Answer/Explanation

Ans:B

Example5

  • $
    \int 2 \tan x \ln (\cos x) d x=
    $

(A) $\cos x[\ln (\cos x)]+C$

(B) $\sin x[\ln (\cos x)]+C$

(C) $-[\ln (\cos x)]^2+C$

(D) $[\ln (\sin x)]^2+C$

▶️Answer/Explanation

Ans:D

Example6

  • $
    \int 2 \tan x \ln (\cos x) d x=
    $

(A) $\cos x[\ln (\cos x)]+C$

(B) $\sin x[\ln (\cos x)]+C$

(C) $-[\ln (\cos x)]^2+C$

(D) $[\ln (\sin x)]^2+C$

▶️Answer/Explanation

Ans:C

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