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

Card 0 of 20

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

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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$ .

Integrate:

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

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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}$x}{4})+C$

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

Tap to see back →

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$ .

Integrate:

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

Tap to see back →

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}$x}{4})+C$

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

Tap to see back →

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$ .

Integrate:

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

Tap to see back →

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}$x}{4})+C$

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

Tap to see back →

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$ .

Integrate:

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

Tap to see back →

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}$x}{4})+C$

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

Tap to see back →

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$ .

Integrate:

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

Tap to see back →

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}$x}{4})+C$

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

Tap to see back →

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$ .

Integrate:

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

Tap to see back →

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}$x}{4})+C$

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

Tap to see back →

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$ .

Integrate:

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

Tap to see back →

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}$x}{4})+C$

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

Tap to see back →

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$ .

Integrate:

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

Tap to see back →

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}$x}{4})+C$

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

Tap to see back →

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$ .

Integrate:

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

Tap to see back →

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}$x}{4})+C$

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

Tap to see back →

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$ .

Integrate:

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

Tap to see back →

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}$x}{4})+C$