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Calculus 2 : Indefinite Integrals

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Example Questions

Example Question #2588 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{67}{(7x)}- 27x^{13})dx\end{align*}$

Possible Answers:

$\frac{(300763ln(x))}{343}-\frac{ (4563x^{14})}{14}$

$\frac{(67ln(x))}{7}-\frac{ (27x^{14})}{14}$

$\frac{(300763ln(x))}{343}-\frac{ (4563x^{14})}{14}+C$

$\frac{(67ln(x))}{7}-\frac{ (27x^{14})}{14}+C$

Correct answer:

$\frac{(67ln(x))}{7}-\frac{ (27x^{14})}{14}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{67}{(7x)}\text{ is }\frac{67}{7}\\&\text{The }a\text{ value for }27x^{13}\text{ is }13\\&\int(\frac{67}{(7x)}- 27x^{13})dx=\frac{(67ln(x))}{7}-\frac{ (27x^{14})}{14}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2589 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(23e^{(19x)})}{5}- 5x^{6})dx\end{align*}$

Possible Answers:

$\frac{(437e^{(19x)})}{5}-\frac{ (180x^{7})}{7}+C$

$\frac{(23e^{(19x)})}{95}-\frac{ (5x^{7})}{7}+C$

$\frac{(437e^{(19x)})}{5}-\frac{ (180x^{7})}{7}$

$\frac{(23e^{(19x)})}{95}-\frac{ (5x^{7})}{7}$

Correct answer:

$\frac{(23e^{(19x)})}{95}-\frac{ (5x^{7})}{7}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{(23e^{(19x)})}{5}\text{ is }19\\&\text{The }a\text{ value for }5x^{6}\text{ is }6\\&\int(\frac{(23e^{(19x)})}{5}- 5x^{6})dx=\frac{(23e^{(19x)})}{95}-\frac{ (5x^{7})}{7}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2590 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(23e^{(15x)} -\frac{ (17\cdot 3^{(20x)})}{4})dx\end{align*}$

Possible Answers:

$345e^{(15x)} -\frac{ (85\cdot 3^{(20x)})}{ln(3)}+C$

$\frac{(23e^{(15x)})}{15}-\frac{ (17\cdot 3^{(20x)})}{(80ln(3))}+C$

$\frac{(23e^{(15x)})}{15}-\frac{ (17\cdot 3^{(20x)})}{(80ln(3))}$

$345e^{(15x)} -\frac{ (85\cdot 3^{(20x)})}{ln(3)}$

Correct answer:

$\frac{(23e^{(15x)})}{15}-\frac{ (17\cdot 3^{(20x)})}{(80ln(3))}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{The }a\text{ value for }23e^{(15x)}\text{ is }15\\&\text{The }a\text{ value for }\frac{(17\cdot 3^{(20x)})}{4}\text{ is }20\\&\int(23e^{(15x)} -\frac{ (17\cdot 3^{(20x)})}{4})dx=\frac{(23e^{(15x)})}{15}-\frac{ (17\cdot 3^{(20x)})}{(80ln(3))}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2591 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(11e^{(13x)})}{40}-\frac{ (45cos(x + 2))}{11})dx\end{align*}$

Possible Answers:

$\frac{(143e^{(13x)})}{40}-\frac{ (180sin(x + 2))}{11}+C$

$\frac{(11e^{(13x)})}{520}-\frac{ (45sin(x + 2))}{11}+C$

$\frac{(11e^{(13x)})}{520}-\frac{ (45sin(x + 2))}{11}$

$\frac{(143e^{(13x)})}{40}-\frac{ (180sin(x + 2))}{11}$

Correct answer:

$\frac{(11e^{(13x)})}{520}-\frac{ (45sin(x + 2))}{11}+C$

Explanation:

$\begin{align*}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{The }a\text{ value for }\frac{(11e^{(13x)})}{40}\text{ is }13\\&\text{In deriving the inside of}\frac{(45cos(x + 2))}{11}\text{, the lone constant goes to zero.}\\&\int(\frac{(11e^{(13x)})}{40}-\frac{ (45cos(x + 2))}{11})dx=\frac{(11e^{(13x)})}{520}-\frac{ (45sin(x + 2))}{11}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2592 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(29\cdot 2^{(\frac{x}{7})})}{10}-\frac{ (59x^{16})}{3})dx\end{align*}$

Possible Answers:

$\frac{(203\cdot 2^{(\frac{x}{7})})}{(10ln(2))}-\frac{ (59x^{17})}{51}+C$

$\frac{(29\cdot 2^{(\frac{x}{7})})}{(70ln(2))}-\frac{ (15104x^{17})}{51}+C$

$\frac{(203\cdot 2^{(\frac{x}{7})})}{(10ln(2))}-\frac{ (59x^{17})}{51}$

$\frac{(29\cdot 2^{(\frac{x}{7})})}{(70ln(2))}-\frac{ (15104x^{17})}{51}$

Correct answer:

$\frac{(203\cdot 2^{(\frac{x}{7})})}{(10ln(2))}-\frac{ (59x^{17})}{51}+C$

Explanation:

$\begin{align*}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{(29\cdot 2^{(\frac{x}{7})})}{10}\text{ is }\frac{1}{7}\\&\text{The }a\text{ value for }\frac{(59x^{16})}{3}\text{ is }16\\&\int(\frac{(29\cdot 2^{(\frac{x}{7})})}{10}-\frac{ (59x^{16})}{3})dx=\frac{(203\cdot 2^{(\frac{x}{7})})}{(10ln(2))}-\frac{ (59x^{17})}{51}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2593 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(2x^{10})}{17}- 8cos(13x))dx\end{align*}$

Possible Answers:

$\frac{(200x^{11})}{187}- 104sin(13x)$

$\frac{(2x^{11})}{187}-\frac{ (8sin(13x))}{13}$

$\frac{(200x^{11})}{187}- 104sin(13x)+C$

$\frac{(2x^{11})}{187}-\frac{ (8sin(13x))}{13}+C$

Correct answer:

$\frac{(2x^{11})}{187}-\frac{ (8sin(13x))}{13}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{The }a\text{ value for }\frac{(2x^{10})}{17}\text{ is }10\\&\text{The }a\text{ value for }8cos(13x)\text{ is }13\\&\int(\frac{(2x^{10})}{17}- 8cos(13x))dx=\frac{(2x^{11})}{187}-\frac{ (8sin(13x))}{13}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2594 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{5}{(132x)}-\frac{ (9x^{5})}{2})dx\end{align*}$

Possible Answers:

$\frac{(5ln(x))}{132}-\frac{ (3x^{6})}{4}$

$\frac{(125ln(x))}{2299968}-\frac{ (75x^{6})}{4}+C$

$\frac{(125ln(x))}{2299968}-\frac{ (75x^{6})}{4}$

$\frac{(5ln(x))}{132}-\frac{ (3x^{6})}{4}+C$

Correct answer:

$\frac{(5ln(x))}{132}-\frac{ (3x^{6})}{4}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{5}{(132x)}\text{ is }\frac{5}{132}\\&\text{The }a\text{ value for }\frac{(9x^{5})}{2}\text{ is }5\\&\int(\frac{5}{(132x)}-\frac{ (9x^{5})}{2})dx=\frac{(5ln(x))}{132}-\frac{ (3x^{6})}{4}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2595 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{cos(x + 20)}{2}-\frac{ (23x^{18})}{6})dx\end{align*}$

Possible Answers:

$200sin(x + 20) -\frac{ (1242x^{19})}{19}+C$

$\frac{sin(x + 20)}{2}-\frac{ (23x^{19})}{114}+C$

$200sin(x + 20) -\frac{ (1242x^{19})}{19}$

$\frac{sin(x + 20)}{2}-\frac{ (23x^{19})}{114}$

Correct answer:

$\frac{sin(x + 20)}{2}-\frac{ (23x^{19})}{114}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{In deriving the inside of}\frac{cos(x + 20)}{2}\text{, the lone constant goes to zero.}\\&\text{The }a\text{ value for }\frac{(23x^{18})}{6}\text{ is }18\\&\int(\frac{cos(x + 20)}{2}-\frac{ (23x^{18})}{6})dx=\frac{sin(x + 20)}{2}-\frac{ (23x^{19})}{114}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2596 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(31x^{6})}{3}-\frac{ 2}{x})dx\end{align*}$

Possible Answers:

$\frac{(31x^{7})}{21}- 2ln(x)+C$

$\frac{(372x^{7})}{7}- 8ln(x)+C$

$\frac{(31x^{7})}{21}- 2ln(x)$

$\frac{(372x^{7})}{7}- 8ln(x)$

Correct answer:

$\frac{(31x^{7})}{21}- 2ln(x)+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\text{The }a\text{ value for }\frac{(31x^{6})}{3}\text{ is }6\\&\text{The }a\text{ value for }\frac{2}{x}\text{ is }2\\&\int(\frac{(31x^{6})}{3}-\frac{ 2}{x})dx=\frac{(31x^{7})}{21}- 2ln(x)+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2597 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{4}{(39x^{15})}- 80cos(5x))dx\end{align*}$

Possible Answers:

$- 16sin(5x) -\frac{ 2}{(273x^{14})}+C$

$- 400sin(5x) -\frac{ 150}{(91x^{14})}+C$

$- 16sin(5x) -\frac{ 2}{(273x^{14})}$

$- 400sin(5x) -\frac{ 150}{(91x^{14})}$

Correct answer:

$- 16sin(5x) -\frac{ 2}{(273x^{14})}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{The }a\text{ value for }\frac{4}{(39x^{15})}\text{ is }-15\\&\text{The }a\text{ value for }80cos(5x)\text{ is }5\\&\int(\frac{4}{(39x^{15})}- 80cos(5x))dx=- 16sin(5x) -\frac{ 2}{(273x^{14})}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2588 : Calculus Ii

Example Question #2589 : Calculus Ii

Example Question #2590 : Calculus Ii

Example Question #2591 : Calculus Ii

Example Question #2592 : Calculus Ii

Example Question #2593 : Calculus Ii

Example Question #2594 : Calculus Ii

Example Question #2595 : Calculus Ii

Example Question #2596 : Calculus Ii

Example Question #2597 : Calculus Ii

All Calculus 2 Resources

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