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

Card 0 of 20

Question

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

Answer

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

Compare your answer with the correct one above

Question

Integrate:

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Integrate:

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Integrate:

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Integrate:

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Integrate:

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Integrate:

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Integrate:

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Integrate:

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Integrate:

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Integrate:

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

Answer

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:

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

Compare your answer with the correct one above

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

In order to evaluate this integral, we will need to use partial fraction decomposition. Multiply both sides of the equation by the common denominator, which is This means that must equal 1, and The answer is .

Integrate:

To integrate, we must first make the following substitution: Rewriting the integral in terms of u and integrating, we get The following rule was used to integrate: Finally, we replace u with our original x term:

In order to evaluate this integral, we will need to use partial fraction decomposition. Multiply both sides of the equation by the common denominator, which is This means that must equal 1, and The answer is .

Integrate:

To integrate, we must first make the following substitution: Rewriting the integral in terms of u and integrating, we get The following rule was used to integrate: Finally, we replace u with our original x term:

In order to evaluate this integral, we will need to use partial fraction decomposition. Multiply both sides of the equation by the common denominator, which is This means that must equal 1, and The answer is .

Integrate:

To integrate, we must first make the following substitution: Rewriting the integral in terms of u and integrating, we get The following rule was used to integrate: Finally, we replace u with our original x term:

In order to evaluate this integral, we will need to use partial fraction decomposition. Multiply both sides of the equation by the common denominator, which is This means that must equal 1, and The answer is .

Integrate:

To integrate, we must first make the following substitution: Rewriting the integral in terms of u and integrating, we get The following rule was used to integrate: Finally, we replace u with our original x term:

In order to evaluate this integral, we will need to use partial fraction decomposition. Multiply both sides of the equation by the common denominator, which is This means that must equal 1, and The answer is .

Integrate:

To integrate, we must first make the following substitution: Rewriting the integral in terms of u and integrating, we get The following rule was used to integrate: Finally, we replace u with our original x term:

In order to evaluate this integral, we will need to use partial fraction decomposition. Multiply both sides of the equation by the common denominator, which is This means that must equal 1, and The answer is .

Integrate:

To integrate, we must first make the following substitution: Rewriting the integral in terms of u and integrating, we get The following rule was used to integrate: Finally, we replace u with our original x term:

In order to evaluate this integral, we will need to use partial fraction decomposition. Multiply both sides of the equation by the common denominator, which is This means that must equal 1, and The answer is .

Integrate:

To integrate, we must first make the following substitution: Rewriting the integral in terms of u and integrating, we get The following rule was used to integrate: Finally, we replace u with our original x term:

In order to evaluate this integral, we will need to use partial fraction decomposition. Multiply both sides of the equation by the common denominator, which is This means that must equal 1, and The answer is .

Integrate:

To integrate, we must first make the following substitution: Rewriting the integral in terms of u and integrating, we get The following rule was used to integrate: Finally, we replace u with our original x term:

In order to evaluate this integral, we will need to use partial fraction decomposition. Multiply both sides of the equation by the common denominator, which is This means that must equal 1, and The answer is .

Integrate:

To integrate, we must first make the following substitution: Rewriting the integral in terms of u and integrating, we get The following rule was used to integrate: Finally, we replace u with our original x term:

In order to evaluate this integral, we will need to use partial fraction decomposition. Multiply both sides of the equation by the common denominator, which is This means that must equal 1, and The answer is .

Integrate:

To integrate, we must first make the following substitution: Rewriting the integral in terms of u and integrating, we get The following rule was used to integrate: Finally, we replace u with our original x term:

Integrate:

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

Integrate:

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

Integrate:

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

Integrate:

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

Integrate:

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

Integrate:

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

Integrate:

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

Integrate:

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

Integrate:

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

Integrate:

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