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Calculus 2 : Indefinite Integrals

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Example Questions

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 #101 : Indefinite Integrals

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

Possible Answers:

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

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

$cot^{-1}x+C$

ln(x^2+1)+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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Example Question #731 : Integrals

$\int \frac{x^2 + x+3}{x^2}dx$

Possible Answers:

$x+ln|x|+\frac{3}{x^2}+C$

$x+ln|x|-\frac{3}{x^2}+C$

$x+ln|x|-\frac{3}{x}+C$

$x+ln|x|+\frac{3}{x}+C$

$ln|x|-\frac{3}{x}+C$

Correct answer:

$x+ln|x|-\frac{3}{x}+C$

Explanation:

We first need to simplify the expression by dividing the numerator by the denominator.

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

Now, we integrate each part. Since this is an indefinite integral, we need to add a constant to the ending.

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

$= x+ln|x|-\frac{3}{x}+C.$

Remember to rewrite the last term to make use of the power rule. $\frac{3}{x^2}=3x^{-2}.$

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Example Question #732 : Integrals

Integrate:

$\int 10t\sin(4t)dt$

Possible Answers:

$-\frac{5t}{2}\cos(4t)-\frac{5}{4}\sin(4t)+C$

$-\frac{5t}{2}\cos(4t)+\frac{5}{8}\sin(4t)+C$

$-\frac{5t}{2}\cos(4t)-\frac{5}{8}\sin(4t)$

$-\frac{5t}{2}\cos(4t)-\frac{5}{8}\sin(4t)+C$

Correct answer:

$-\frac{5t}{2}\cos(4t)-\frac{5}{8}\sin(4t)+C$

Explanation:

To integrate, we must use integration by parts, which is given by the following:

$\int udv=uv-\int vdu$

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

u=10t , du=10dt

$dv=\sin(4t)$ , $v=-\frac{\cos(4t)}{4}$

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

The rules used are as follows:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\int \sin(x)dx=-\cos(x)+C$

Now, use the above formula to rewrite the integral:

$\frac{-5t\cos(4t)}{2}-\frac{5}{2}\int \cos(4t)dt$

Next, we integrate to get our final answer:

$-\frac{5t}{2}\cos(4t)-\frac{5}{8}\sin(4t)+C$

The integration was performed using the following rule:

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

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Example Question #733 : Integrals

$\int 4x^2-1dx$

Possible Answers:

$\frac{4x^3}{3}-x+C$

$\frac{x^3}{3}-x+C$

4x^3-x+C

$\frac{4x^3}{3}-3x+C$

$\frac{4x^3}{3}-x$

Correct answer:

$\frac{4x^3}{3}-x+C$

Explanation:

Remember that when integrating, raise the exponent by 1 and then put that result on the denominator. A constant yields an integral of x. Therefore, your answer is: $\frac{4x^3}{3}-x$ . Remember to add a +C because it is an indefinite integral: $\frac{4x^3}{3}-x+C$ .

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Example Question #734 : Integrals

$\int 4x^{\frac{1}{3}}-2xdx$

Possible Answers:

$3x^{\frac{4}{3}}-x^2$

$3x^{\frac{4}{3}}-x+C$

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

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

$3x^{\frac{1}{3}}-x^2+C$

Correct answer:

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

Explanation:

To integrate, remember that you add one to the exponent and then also put that result on the denominator. After integrating, you get: $4(\frac{x^{\frac{4}{3}}}{\frac{4}{3}})-\frac{2x^2}{2}$ . Simplify and get $3x^{\frac{4}{3}}-x^2$ . Then, remember to add +C because it is an indefinite integral: $3x^{\frac{4}{3}}-x^2+C.$

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Example Question #735 : Integrals

$\int (4x-1)(2x+3)dx$

Possible Answers:

$\frac{8}{3}x^3+5x^2-3x+C$

$\frac{8}{3}x^3-5x^2-3x+C$

$\frac{8}{3}x^3+5x^2-x+C$

$\frac{8}{3}x^3+5x^2-3x$

x^3+5x^2-3x+C

Correct answer:

$\frac{8}{3}x^3+5x^2-3x+C$

Explanation:

First off, multiply the binomials using FOIL: $\int 8x^2+10x-3dx$ . Then, integrate, remembering to raise the exponent by 1 and then putting that result on the denominator: $\frac{8x^3}{3}+\frac{10x^2}{2}-3x$ . Simplify and add a +C because it is an indefinite integral: $\frac{8}{3}x^3+5x^2-3x+C$ .

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Example Question #102 : Indefinite Integrals

$\int_{1}^{3}\frac{4}{x}dx$

Possible Answers:

2.3944

4.3444

1.0264

4.3944

Correct answer:

4.3944

Explanation:

Recall that when integrating a single variable on the denominator, take ln of that term. Therefore, the integration is $4ln\left | x \right |$ . Then, evaluate at 3 and then 1. Subtract the results: 4ln3-4ln1=4.3944 .

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Example Question #737 : Integrals

$\int (2-x)(3+x^2)dx$

Possible Answers:

$-\frac{x^4}{4}+\frac{2x^3}{3}-\frac{3x^2}{2}+6x$

$-\frac{x^4}{4}+\frac{2x^3}{3}+\frac{3x^2}{2}+6x+C$

$-\frac{x^4}{4}+\frac{2x^3}{3}-\frac{3x^2}{2}+6x+C$

$\frac{x^4}{4}+\frac{2x^3}{3}-\frac{3x^2}{2}+6x+C$

Correct answer:

$-\frac{x^4}{4}+\frac{2x^3}{3}-\frac{3x^2}{2}+6x+C$

Explanation:

The first step here should be multiplying the binomials:

(2-x)(3+x^2)=-x^3+2x^2-3x+6 .

Then, integrate. Remember to raise each power by one and then also put that result on the denominator.

Add a +C at the end because it is an indefinite integral.

Therefore, your answer should like this:

$-\frac{x^4}{4}+\frac{2x^3}{3}-\frac{3x^2}{2}+6x+C$ .

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Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #101 : Indefinite Integrals

Example Question #372 : Finding Integrals

Example Question #101 : Indefinite Integrals

Example Question #731 : Integrals

Example Question #732 : Integrals

Example Question #733 : Integrals

Example Question #734 : Integrals

Example Question #735 : Integrals

Example Question #102 : Indefinite Integrals

Example Question #737 : Integrals

All Calculus 2 Resources

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