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Calculus 2 : Indefinite Integrals

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Example Questions

Example Question #471 : Finding Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(68cos(x + 9))}{7}+\frac{ 23}{(2x^{7})})dx\end{align*}$

Possible Answers:

$\frac{(68sin(x + 9))}{7}-\frac{ 23}{(12x^{6})}$

$\frac{(68sin(x + 9))}{7}-\frac{ 23}{(12x^{6})}+C$

$\frac{(5508sin(x + 9))}{7}-\frac{ 1127}{(12x^{6})}+C$

$\frac{(5508sin(x + 9))}{7}-\frac{ 1127}{(12x^{6})}$

Correct answer:

$\frac{(68sin(x + 9))}{7}-\frac{ 23}{(12x^{6})}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{In deriving the inside of}\frac{(68cos(x + 9))}{7}\text{, the lone constant goes to zero.}\\&\text{The }a\text{ value for }\frac{23}{(2x^{7})}\text{ is }-7\\&\int(\frac{(68cos(x + 9))}{7}+\frac{ 23}{(2x^{7})})dx=\frac{(68sin(x + 9))}{7}-\frac{ 23}{(12x^{6})}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #828 : Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{sin(6x)}{50}+\frac{ x^{8}}{29})dx\end{align*}$

Possible Answers:

$\frac{(64x^{9})}{261}-\frac{ (3cos(6x))}{25}$

$\frac{(64x^{9})}{261}-\frac{ (3cos(6x))}{25}+C$

$\frac{x^{9}}{261}-\frac{ cos(6x)}{300}$

$\frac{x^{9}}{261}-\frac{ cos(6x)}{300}+C$

Correct answer:

$\frac{x^{9}}{261}-\frac{ cos(6x)}{300}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{The }a\text{ value for }\frac{x^{8}}{29}\text{ is }8\\&\text{The }a\text{ value for }\frac{sin(6x)}{50}\text{ is }6\\&\int(\frac{sin(6x)}{50}+\frac{ x^{8}}{29})dx=\frac{x^{9}}{261}-\frac{ cos(6x)}{300}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #829 : Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(2cos(4x))}{7}+\frac{ (37e^{(15x)})}{8})dx\end{align*}$

Possible Answers:

$\frac{(555e^{(15x)})}{8}+\frac{ (8sin(4x))}{7}+C$

$\frac{(37e^{(15x)})}{120}+\frac{ sin(4x)}{14}$

$\frac{(37e^{(15x)})}{120}+\frac{ sin(4x)}{14}+C$

$\frac{(555e^{(15x)})}{8}+\frac{ (8sin(4x))}{7}$

Correct answer:

$\frac{(37e^{(15x)})}{120}+\frac{ sin(4x)}{14}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\text{The }a\text{ value for }\frac{(2cos(4x))}{7}\text{ is }4\\&\text{The }a\text{ value for }\frac{(37e^{(15x)})}{8}\text{ is }15\\&\int(\frac{(2cos(4x))}{7}+\frac{ (37e^{(15x)})}{8})dx=\frac{(37e^{(15x)})}{120}+\frac{ sin(4x)}{14}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2581 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{sin(x + 6)}{34}+\frac{ 1}{(11x)})dx\end{align*}$

Possible Answers:

$\frac{ln(x)}{11}-\frac{ cos(x + 6)}{34}$

$\frac{ln(x)}{1331}-\frac{ (18cos(x + 6))}{17}+C$

$\frac{ln(x)}{11}-\frac{ cos(x + 6)}{34}+C$

$\frac{ln(x)}{1331}-\frac{ (18cos(x + 6))}{17}$

Correct answer:

$\frac{ln(x)}{11}-\frac{ cos(x + 6)}{34}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\text{In deriving the inside of}\frac{sin(x + 6)}{34}\text{, the lone constant goes to zero.}\\&\text{The }a\text{ value for }\frac{1}{(11x)}\text{ is }\frac{1}{11}\\&\int(\frac{sin(x + 6)}{34}+\frac{ 1}{(11x)})dx=\frac{ln(x)}{11}-\frac{ cos(x + 6)}{34}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2582 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(3e^{(10x)})}{16}+\frac{ 1}{(116x)})dx\end{align*}$

Possible Answers:

$\frac{(3e^{(10x)})}{160}+\frac{ ln(x)}{116}$

$\frac{(15e^{(10x)})}{8}+\frac{ ln(x)}{1560896}+C$

$\frac{(3e^{(10x)})}{160}+\frac{ ln(x)}{116}+C$

$\frac{(15e^{(10x)})}{8}+\frac{ ln(x)}{1560896}$

Correct answer:

$\frac{(3e^{(10x)})}{160}+\frac{ ln(x)}{116}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\text{The }a\text{ value for }\frac{1}{(116x)}\text{ is }\frac{1}{116}\\&\text{The }a\text{ value for }\frac{(3e^{(10x)})}{16}\text{ is }10\\&\int(\frac{(3e^{(10x)})}{16}+\frac{ 1}{(116x)})dx=\frac{(3e^{(10x)})}{160}+\frac{ ln(x)}{116}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2583 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(58\cdot 2^{(\frac{x}{20})} +\frac{ x^{8}}{46})dx\end{align*}$

Possible Answers:

$\frac{(1160\cdot 2^{(\frac{x}{20})})}{ln(2)}+\frac{ x^{9}}{414}+C$

$\frac{(1160\cdot 2^{(\frac{x}{20})})}{ln(2)}+\frac{ x^{9}}{414}$

$\frac{(29\cdot 2^{(\frac{x}{20})})}{(10ln(2))}+\frac{ (32x^{9})}{207}+C$

$\frac{(29\cdot 2^{(\frac{x}{20})})}{(10ln(2))}+\frac{ (32x^{9})}{207}$

Correct answer:

$\frac{(1160\cdot 2^{(\frac{x}{20})})}{ln(2)}+\frac{ x^{9}}{414}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }58\cdot 2^{(\frac{x}{20})}\text{ is }\frac{1}{20}\\&\text{The }a\text{ value for }\frac{x^{8}}{46}\text{ is }8\\&\int(58\cdot 2^{(\frac{x}{20})} +\frac{ x^{8}}{46})dx=\frac{(1160\cdot 2^{(\frac{x}{20})})}{ln(2)}+\frac{ x^{9}}{414}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2584 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{cos(x + 14)}{47}+\frac{ (30e^{(2x)})}{7})dx\end{align*}$

Possible Answers:

$\frac{(15e^{(2x)})}{7}+\frac{ sin(x + 14)}{47}+C$

$\frac{(60e^{(2x)})}{7}+\frac{ (196sin(x + 14))}{47}$

$\frac{(15e^{(2x)})}{7}+\frac{ sin(x + 14)}{47}$

$\frac{(60e^{(2x)})}{7}+\frac{ (196sin(x + 14))}{47}+C$

Correct answer:

$\frac{(15e^{(2x)})}{7}+\frac{ sin(x + 14)}{47}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{The }a\text{ value for }\frac{(30e^{(2x)})}{7}\text{ is }2\\&\text{In deriving the inside of}\frac{cos(x + 14)}{47}\text{, the lone constant goes to zero.}\\&\int(\frac{cos(x + 14)}{47}+\frac{ (30e^{(2x)})}{7})dx=\frac{(15e^{(2x)})}{7}+\frac{ sin(x + 14)}{47}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2585 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{e^{(8x)}}{7}+\frac{ (7\cdot 2^{(\frac{x}{9})})}{2})dx\end{align*}$

Possible Answers:

$\frac{(8e^{(8x)})}{7}+\frac{ (7\cdot 2^{(\frac{x}{9})})}{(18ln(2))}$

$\frac{(8e^{(8x)})}{7}+\frac{ (7\cdot 2^{(\frac{x}{9})})}{(18ln(2))}+C$

$\frac{e^{(8x)}}{56}+\frac{ (63\cdot 2^{(\frac{x}{9})})}{(2ln(2))}$

$\frac{e^{(8x)}}{56}+\frac{ (63\cdot 2^{(\frac{x}{9})})}{(2ln(2))}+C$

Correct answer:

$\frac{e^{(8x)}}{56}+\frac{ (63\cdot 2^{(\frac{x}{9})})}{(2ln(2))}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\text{The }a\text{ value for }\frac{(7\cdot 2^{(\frac{x}{9})})}{2}\text{ is }\frac{1}{9}\\&\text{The }a\text{ value for }\frac{e^{(8x)}}{7}\text{ is }8\\&\int(\frac{e^{(8x)}}{7}+\frac{ (7\cdot 2^{(\frac{x}{9})})}{2})dx=\frac{e^{(8x)}}{56}+\frac{ (63\cdot 2^{(\frac{x}{9})})}{(2ln(2))}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2586 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(3sin(19x))}{5}+\frac{ (34\cdot 3^{(\frac{x}{11})})}{3})dx\end{align*}$

Possible Answers:

$\frac{(374\cdot 3^{(\frac{x}{11})})}{(3ln(3))}-\frac{ (3cos(19x))}{95}+C$

$\frac{(34\cdot 3^{(\frac{x}{11})})}{(33ln(3))}-\frac{ (57cos(19x))}{5}+C$

$\frac{(374\cdot 3^{(\frac{x}{11})})}{(3ln(3))}-\frac{ (3cos(19x))}{95}$

$\frac{(34\cdot 3^{(\frac{x}{11})})}{(33ln(3))}-\frac{ (57cos(19x))}{5}$

Correct answer:

$\frac{(374\cdot 3^{(\frac{x}{11})})}{(3ln(3))}-\frac{ (3cos(19x))}{95}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{The }a\text{ value for }\frac{(34\cdot 3^{(\frac{x}{11})})}{3}\text{ is }\frac{1}{11}\\&\text{The }a\text{ value for }\frac{(3sin(19x))}{5}\text{ is }19\\&\int(\frac{(3sin(19x))}{5}+\frac{ (34\cdot 3^{(\frac{x}{11})})}{3})dx=\frac{(374\cdot 3^{(\frac{x}{11})})}{(3ln(3))}-\frac{ (3cos(19x))}{95}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2587 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(28x^{12})}{5}-\frac{ x^{8}}{5})dx\end{align*}$

Possible Answers:

$\frac{(x^{9}\cdot (252x^{4} - 13))}{585}$

$\frac{(4032x^{13})}{65}-\frac{ (64x^{9})}{45}+C$

$\frac{(x^{9}\cdot (252x^{4} - 13))}{585}+C$

$\frac{(4032x^{13})}{65}-\frac{ (64x^{9})}{45}$

Correct answer:

$\frac{(x^{9}\cdot (252x^{4} - 13))}{585}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{(28x^{12})}{5}\text{ is }12\\&\text{The }a\text{ value for }\frac{x^{8}}{5}\text{ is }8\\&\int(\frac{(28x^{12})}{5}-\frac{ x^{8}}{5})dx=\frac{(x^{9}\cdot (252x^{4} - 13))}{585}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #471 : Finding Integrals

Example Question #828 : Integrals

Example Question #829 : Integrals

Example Question #2581 : Calculus Ii

Example Question #2582 : Calculus Ii

Example Question #2583 : Calculus Ii

Example Question #2584 : Calculus Ii

Example Question #2585 : Calculus Ii

Example Question #2586 : Calculus Ii

Example Question #2587 : Calculus Ii

All Calculus 2 Resources

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