Substitution of Variables (by parts and simple partial fractions)

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

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1

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

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

CORRECT

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

0

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

0

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

0

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

0

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

This means that must equal 1, and

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

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

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

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

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

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

2

Integrate:

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

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

CORRECT

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

0

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

0

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

0

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

0

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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