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Calculus 2 : Integrals

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Example Questions

Example Question #721 : Integrals

Integrate:

$\int \frac{\ln x}{x^3}dx$

Possible Answers:

$\frac{-\ln x-1}{2x^2}+C$

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

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

$-\frac{\ln x}{2x^2}-\frac{1}{4x^2}$

Correct answer:

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

Explanation:

To integrate, we must integrate by parts, which uses the following formula:

$\int u dv=uv-\int v du$

We must choose our u and dv, and differentiate and integrate, respectively:

$u=\ln(x)$ , $du=\frac{dx}{x}$

$dv= x^{-3}, v=-\frac{1}{2}x^{-2}$

The following rules were used:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln x=\frac{1}{x}$ , $\int x^ndx=\frac{x^{n+1}}{n+1}+C$

Note that we do not include the constant of integration in this step.

Now, using the above formula, rewrite the integral:

$(\ln x)(-\frac{1}{2}x^{-2})+\int \frac{1}{2}x^{-2}\frac{1}{x}dx=-\frac{1}{2}\ln(x)x^{-2} +\frac{1}{2}\int x^{-3}dx$

After integrating, we get our final answer

$-\frac{1}{2}\ln(x)x^{-2} +\frac{1}{2}\cdot \frac{x^{-2}}{2}=-\frac{\ln(x)}{2x^2}-\frac{1}{4x^2}+C$

using the same integration rule as above.

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Example Question #722 : Integrals

Integrate:

$\int \ln(\sqrt[4]{x})dx$

Possible Answers:

$\frac{1}{4}(x \ln x -x)+C$

$x \ln x -x+C$

$\frac{x \ln x}{4} -x+C$

$\frac{1}{4}(x \ln x -x)$

Correct answer:

$\frac{1}{4}(x \ln x -x)+C$

Explanation:

To integrate, we must integrate by parts, which is done using the following formula:

$\int udv=uv-\int vdu$

Before rewriting the integrand using integration by parts, we can use a logarithm property to rewrite it as

$\int \ln x^{\frac{1}{4}} dx=\frac{1}{4} \int \ln x dx$

Now, we must designate our u and dv, and differentiate and integrate, respectively:

$u=\ln x$ , $du= \frac{dx}{x}$

dv=dx , v=x

We used the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$ , $\int dx =x +C$

Note that we don't include the constant of integration in this step.

Using the above formula, we can rewrite the integral as

$\frac{1}{4}(x\ln x - \int x \cdot \frac{dx}{x})=\frac{1}{4}(x\ln x -\int dx)$

Integrating, we get

$\frac{1}{4}(x\ln x - x+C)$

using the same rule as above.

Our final answer is

$\frac{1}{4}(x \ln x -x)+C$

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Example Question #723 : Integrals

Integrate:

$\int \sin^2(\frac{3x}{4})dx$

Possible Answers:

$\frac{1}{2}x-\frac{1}{3}\cos(\frac{3}{2}x)+C$

$\frac{1}{2}x+\frac{1}{3}\cos(\frac{3}{2}x)+C$

$x+\frac{1}{3}\cos(\frac{3}{2}x)+C$

$\frac{1}{2}x+\frac{1}{3}\sin(\frac{3}{2}x)+C$

$\frac{1}{2}x+\frac{1}{3}\cos(\frac{3}{2}x)$

Correct answer:

$\frac{1}{2}x+\frac{1}{3}\cos(\frac{3}{2}x)+C$

Explanation:

To integrate, we must rewrite the integrand using the half angle identity:

$\int \frac{1}{2}(1-\sin(\frac{3}{2}x))dx$

Now, integrate:

$\int \frac{1}{2}(1-\sin(\frac{3}{2}x))dx=\frac{1}{2}(x+\frac{1}{3}\cos(\frac{3}{2}x))=\frac{1}{2}x+\frac{1}{3}\cos(\frac{3}{2}x)+C$

We used the following rules to integrate:

$\int dx=x+C$ , $\int \sin(x)dx=-\cos(x)+C$

Note that we combined the two constants of integration to make one C.

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Example Question #371 : Finding Integrals

Evaluate.

$\int 12 + x \ dx$

Possible Answers:

Answer not listed.

F(x) = 1 +C

$F(x) = \frac{6x}{2} + \frac{x^2}{2} +C$

F(x) = 12x + 2x +C

$F(x) = 12x + \frac{x^2}{2} + C$

Correct answer:

$F(x) = 12x + \frac{x^2}{2} + C$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, f(x) = 12 + x .

The antiderivative is $F(x) = 12x + \frac{x^2}{2} +C$ .

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Example Question #95 : Indefinite Integrals

Evaluate.

$\int \frac{1}{2}x - 30x^{14} \ dx$

Possible Answers:

$F(x) = \frac{1}{x} - 15x +C$

$F(x) = \frac{x^2}{2} - 30x^{14} +C$

$F(x) = 4x^2- 14x^{13} +C$

Answer not listed.

$F(x) = \frac{x^2}{4} - 2x^{15} +C$

Correct answer:

$F(x) = \frac{x^2}{4} - 2x^{15} +C$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{1}{2}x - 30x^{14}$ .

The antiderivative is $F(x) = \frac{x^2}{4} - 2x^{15} +C$ .

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Example Question #96 : Indefinite Integrals

Evaluate.

$\int \frac{3}{x-5} \ dx$

Possible Answers:

$F(x) = \frac{\ln(x-5)}{3} +C$

$F(x) = \frac{3}{ln(x-5)} +C$

$F(x) = \frac{1}{3}\ln(x-5) +C$

Answer not listed.

$F(x) = \ln(x-5) +C$

Correct answer:

Answer not listed.

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{3}{x-5}$ .

The antiderivative is $F(x) = 3\ln(x-5) +C$ .

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Example Question #97 : Indefinite Integrals

Evaluate.

$\int e^{7x} \ dx$

Possible Answers:

Answer not listed.

$F(x) = \frac{7}{e^{7x}}$

$F(x) =7e^{7x} +C$

$F(x) = \frac{e^{7x}}{7} +C$

$F(x) = e^{7x} +C$

Correct answer:

$F(x) = \frac{e^{7x}}{7} +C$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = e^{7x}$ .

The antiderivative is $F(x) = \frac{e^{7x}}{7} +C$ .

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Example Question #101 : Indefinite Integrals

Evaluate.

$\int \cos(6x) \ dx$

Possible Answers:

$F(x) =- \frac{\sin(6x)}{6} +C$

$F(x) =- 6\sin(6x) +C$

Answer not listed.

$F(x) = \frac{6}{\sin(6x)} +C$

$F(x) = \frac{\sin(6x)}{6} +C$

Correct answer:

$F(x) = \frac{\sin(6x)}{6} +C$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \cos(6x)$ .

The antiderivative is $F(x) = \frac{\sin(6x)}{6} +C$ .

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Example Question #372 : Finding Integrals

Evaluate.

$\int \frac{x-4}{x} \ dx$

Possible Answers:

$F(x) = x\ln(x-4) +C$

Answer not listed.

$F(x) = x\ln(x) +C$

$F(x) =\frac{4}{x-4} +C$

$F(x) = x-4\ln(x) +C$

Correct answer:

$F(x) = x-4\ln(x) +C$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{x-4}{x}$ .

The antiderivative is $F(x) = x-4\ln(x) +C$ .

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Example Question #724 : Integrals

$\int \frac{1}{x^2+1}dx$

Possible Answers:

$\frac{-1}{x}+C$

ln(x^2+1)+C

$cot^{-1}x+C$

$tan^{-1}(x)+C$

Correct answer:

$tan^{-1}(x)+C$

Explanation:

This integral is a definition and should be committed to memory. These functions are extremely useful when integrating.

$\int \frac{1}{x^2+1}dx=tan^{-1}x+C.$

Remember, every time you have an indefinite integral, you need to add on a constant to the end.

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #721 : Integrals

Example Question #722 : Integrals

Example Question #723 : Integrals

Example Question #371 : Finding Integrals

Example Question #95 : Indefinite Integrals

Example Question #96 : Indefinite Integrals

Example Question #97 : Indefinite Integrals

Example Question #101 : Indefinite Integrals

Example Question #372 : Finding Integrals

Example Question #724 : Integrals

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