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AP Calculus BC : Substitution of Variables (by parts and simple partial fractions)

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Example Questions

Example Question #1 : Substitution Of Variables (By Parts And Simple Partial Fractions)

$\int\frac{x-1}{(x-3)(x+4)} dx=$

Possible Answers:

$\frac{5}{7(x+4)}\ln |x|+\frac{2}{7(x-3)}\ln |x|+C$

$\frac{5}{7(x+4)}+\frac{2}{7(x+2)}+C$

$\frac{(x-3)^2}{}7+\frac{2(x+4)^2}{7}+C$

$\frac{1}{7}\ln|x-2|+\frac{1}{7}\ln|x-4|+C$

$\frac{2}{7}\ln|x-3|+\frac{5}{7}\ln| x+4|+C$

Correct answer:

$\frac{2}{7}\ln|x-3|+\frac{5}{7}\ln| x+4|+C$

Explanation:

In order to evaluate this integral, we will need to use partial fraction decomposition.

$\frac{x-1}{(x-3)(x+4)}=\frac{A}{x-3}+\frac{B}{x+4}$

Multiply both sides of the equation by the common denominator, which is (x-3)(x+4).

x-1=A(x+4)+B(x-3)

x-1=Ax+4A+Bx-3B=(A+B)x+4A-3B

This means that A+B must equal 1, and 4A-3B=-1.

A+B=1

A=1-B

4(1-B)-3B=-1

B=5/7, A=2/7

$\frac{x-1}{(x-3)(x+4)}=\frac{2/7}{x-3}+\frac{5/7}{x+4}$

$\int(\frac{2}{7}\cdot \frac{1}{x-3}+\frac{5}{7}\cdot \frac{1}{x+4})dx$

$=\int\frac{2}{7}\cdot\frac{1}{x-3} dx +\int\frac{5}{7}\cdot\frac{1}{x+4}dx$

$=\frac{2}{7}\int\frac{dx}{x-3}+\frac{5}{7}\int\frac{dx}{x+4}$

$=\frac{2}{7}\ln|x-3|+\frac{5}{7}\ln| x+4|+C$

The answer is $\frac{2}{7}\ln|x-3|+\frac{5}{7}\ln| x+4|+C$ .

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Example Question #2 : Substitution Of Variables (By Parts And Simple Partial Fractions)

Integrate:

$\int \frac{dx}{2x^2+16}$

Possible Answers:

$\frac{1}{4\sqrt{2}}\arctan(\frac{x}{4})+C$

$\frac{1}{4\sqrt{2}}\arctan(\frac{\sqrt{2}x}{4})+C$

$\frac{1}{\sqrt{2}}\arctan(\frac{\sqrt{2}x}{4})+C$

$\frac{1}{4\sqrt{2}}\arcsin(\frac{x}{4})+C$

$\arccos(\frac{x}{4})+C$

Correct answer:

$\frac{1}{4\sqrt{2}}\arctan(\frac{\sqrt{2}x}{4})+C$

Explanation:

To integrate, we must first make the following substitution:

$u=\sqrt{2}x, du=\sqrt{2}dx$

Rewriting the integral in terms of u and integrating, we get

$\frac{1}{\sqrt{2}}\int \frac{du}{u^2+16}= \frac{1}{4\sqrt{2}}\arctan(\frac{u}{4})+C$

The following rule was used to integrate:

$\int \frac{du}{u^2+a^2}=\frac{1}{a}\arctan(\frac{x}{a})+C$

Finally, we replace u with our original x term:

$\frac{1}{4\sqrt{2}}\arctan(\frac{\sqrt{2}x}{4})+C$

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AP Calculus BC : Substitution of Variables (by parts and simple partial fractions)

Study concepts, example questions & explanations for AP Calculus BC

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Example Questions

Example Question #1 : Substitution Of Variables (By Parts And Simple Partial Fractions)

Example Question #2 : Substitution Of Variables (By Parts And Simple Partial Fractions)

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