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

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Question

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

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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$ .

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Question

Integrate:

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

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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}$x}{4})+C$

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Question

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

Tap to reveal answer

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$ .

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Question

Integrate:

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

Tap to reveal answer

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}$x}{4})+C$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$ .

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Question

Integrate:

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

Tap to reveal answer

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}$x}{4})+C$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$ .

← Didn't Know|Knew It →

Question

Integrate:

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

Tap to reveal answer

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}$x}{4})+C$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$ .

← Didn't Know|Knew It →

Question

Integrate:

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

Tap to reveal answer

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}$x}{4})+C$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$ .

← Didn't Know|Knew It →

Question

Integrate:

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

Tap to reveal answer

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}$x}{4})+C$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$ .

← Didn't Know|Knew It →

Question

Integrate:

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

Tap to reveal answer

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}$x}{4})+C$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$ .

← Didn't Know|Knew It →

Question

Integrate:

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

Tap to reveal answer

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}$x}{4})+C$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$ .

← Didn't Know|Knew It →

Question

Integrate:

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

Tap to reveal answer

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}$x}{4})+C$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$ .

← Didn't Know|Knew It →

Question

Integrate:

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

Tap to reveal answer

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}$x}{4})+C$

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Knew It

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