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Calculus 2 : Indefinite Integrals

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Example Questions

Example Question #71 : Indefinite Integrals

Evaluate.

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

Possible Answers:

Answer not listed.

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

$F(x) = 7e^{7x} - 7$

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

$F(x) = e^{7x} - 7$

Correct answer:

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

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, $f(x) = e^{7x} + 7$ .

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

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Example Question #72 : Indefinite Integrals

Evaluate.

$\int \sin(x) + 6 - \frac{1}{5x}\ dx$

Possible Answers:

$F(x) = - \cos(x) + 6x - \frac{\ln (x)}{5}$

$F(x) = - \sin(x) + 6x - \frac{\ln (5x)}{5}$

Answer not listed.

$F(x) = - \cos(x) + 6 - \frac{\ln (5x)}{5}$

$F(x) = - \cos(x) + 6x - 5\ln (x)$

Correct answer:

$F(x) = - \cos(x) + 6x - \frac{\ln (x)}{5}$

Explanation:

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, $f(x) = \sin(x) + 6 - \frac{1}{5x}$ .

The antiderivative is $F(x) = - \cos(x) + 6x - \frac{\ln (x)}{5}$ .

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Example Question #71 : Indefinite Integrals

$\int \frac{2x^2-8x+2}{2}dx$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

First, make the fraction three separate terms: $\int x^2-4x+1dx$ . Now, integrate as normal, remembering to raise the exponent by 1 and then also putting that result on the denominator: $\frac{x^3}{3}-2x^2+x$ . Remember to add a C at the end because it is an indefinite integral: $\frac{x^3}{3}-2x^2+x+C$ .

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Example Question #72 : Indefinite Integrals

$\int 3x^4-2x^2+x-9dx$

Possible Answers:

$\frac{3x^5}{5}-\frac{2x^3}{3}+x^2-9x+C$

$\frac{3x^5}{5}-\frac{2x^3}{3}+\frac{x^2}{2}-9x$

$\frac{3x^5}{5}-\frac{2x^3}{3}+\frac{x^2}{2}-9x+C$

$x^5-\frac{2x^3}{3}+\frac{x^2}{2}-9x+C$

$\frac{3x^5}{5}+\frac{2x^3}{3}+\frac{x^2}{2}-9x+C$

Correct answer:

$\frac{3x^5}{5}-\frac{2x^3}{3}+\frac{x^2}{2}-9x+C$

Explanation:

Recall that when integrating, raise the exponent of the x term by 1 and then also put that result on the denominator. Integrate each term separately and get an answer of $\frac{3x^5}{5}-\frac{2x^3}{3}+\frac{x^2}{2}-9x$ . Remember to add a C at the end because it is an indefinite integral: $\frac{3x^5}{5}-\frac{2x^3}{3}+\frac{x^2}{2}-9x+C.$

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Example Question #71 : Indefinite Integrals

$\int x^2-1dx$

Possible Answers:

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

x^3-x+C

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

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

Correct answer:

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

Explanation:

Remember, when integrating, raise the exponent by 1 and then put that result in the denominator as well. Therefore, you should get: $\frac{x^3}{3}-x$ . Remember to add a C at the end to get $\frac{x^3}{3}-x+C$ because it is an indefinite integral.

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Example Question #72 : Indefinite Integrals

Evaluate the following indefinite integral:

$\int (12x^3+14x)dx$

Possible Answers:

36x^4+14x^2+C

3x^4+7x^2+C

18x^4+7x^2+C

36x^2+14+C

Correct answer:

3x^4+7x^2+C

Explanation:

$\int (12x^3+14x)dx=\frac{1}{4}12x^4+\frac{1}{2}14x^2+C=3x^4+7x^2+C$

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Example Question #704 : Integrals

Evaluate.

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

Possible Answers:

$F(x) = x^4 + \frac{\sin (2x)}{2}$

$F(x) = x^4 - \sin (2x)$

$F(x) = \frac{x^3}{4} - \frac{\sin (x)}{2}+C$

Answer not listed

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

Correct answer:

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

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, $f(x) = x^3 - \cos (2x)$ .

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

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Example Question #701 : Integrals

Evaluate.

$\int 5e^{5x} \ dx$

Possible Answers:

Answer not listed

$F(x) = \frac{1}{5}e^{5x}+C$

$F(x) = e^{5x} + x+C$

$F(x) = \frac{e^{5x}}{5}$

$F(x) = e^{5x}+C$

Correct answer:

$F(x) = e^{5x}+C$

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, $f(x) = 5e^{5x}$ .

The antiderivative is $F(x) = e^{5x}+C$ .

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Example Question #2453 : Calculus Ii

Evaluate.

$\int \frac{1}{6x+5} \ dx$

Possible Answers:

$F(x) = \frac{\ln(6x)}{x}$

$F(x) = \frac{\ln(6x+5)}{6}+C$

$F(x) = \ln(6x+5)$

Answer not listed

$F(x) = \frac{\ln(x)}{6x+5}+C$

Correct answer:

$F(x) = \frac{\ln(6x+5)}{6}+C$

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, $f(x) = \frac{1}{6x+5}$ .

The antiderivative is $F(x) = \frac{\ln(6x+5)}{6}+C$ .

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Example Question #702 : Integrals

Evaluate.

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

Possible Answers:

$F(x) = 3 \ln(x) - \sin(3x)+C$

Answer not listed

$F(x) = \frac{\ln(3x) - \sin(3x)}{3}$

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

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

Correct answer:

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

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, $f(x) = -\cos(3x) + \frac{1}{3x}$ .

The antiderivative is $F(x) = \frac{\ln(x) - \sin(3x)}{3}+C$ .

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Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #71 : Indefinite Integrals

Example Question #72 : Indefinite Integrals

Example Question #71 : Indefinite Integrals

Example Question #72 : Indefinite Integrals

Example Question #71 : Indefinite Integrals

Example Question #72 : Indefinite Integrals

Example Question #704 : Integrals

Example Question #701 : Integrals

Example Question #2453 : Calculus Ii

Example Question #702 : Integrals

All Calculus 2 Resources

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