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Calculus 2 : Indefinite Integrals

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Example Questions

Example Question #2598 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(76cos(x + 6))}{3}-\frac{ 25}{(102x)})dx\end{align*}$

Possible Answers:

$\frac{(76sin(x + 6))}{3}-\frac{ (25ln(x))}{102}+C$

$912sin(x + 6) -\frac{ (15625ln(x))}{1061208}+C$

$\frac{(76sin(x + 6))}{3}-\frac{ (25ln(x))}{102}$

$912sin(x + 6) -\frac{ (15625ln(x))}{1061208}$

Correct answer:

$\frac{(76sin(x + 6))}{3}-\frac{ (25ln(x))}{102}+C$

Explanation:

$\begin{align*}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\text{In deriving the inside of}\frac{(76cos(x + 6))}{3}\text{, the lone constant goes to zero.}\\&\text{The }a\text{ value for }\frac{25}{(102x)}\text{ is }\frac{25}{102}\\&\int(\frac{(76cos(x + 6))}{3}-\frac{ 25}{(102x)})dx=\frac{(76sin(x + 6))}{3}-\frac{ (25ln(x))}{102}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2599 : Calculus Ii

Find the solution to the Integral:

$\int \frac{ln^7(x)}{x}dx$

Possible Answers:

8ln(x)+c

8ln^8(x)+c

7ln^6(x)+c

$\frac{ln^8(x)}{8}+c$

$\frac{}1{}8ln^2(x)+c$

Correct answer:

$\frac{ln^8(x)}{8}+c$

Explanation:

To Evaluate this Integral we must use substitution:

Let

u=ln(x) and $du=\frac{1}{x}dx$

Therefore:

$\int \frac{ln^7(x)}{x}dx=\int u^7du$

$\int u^7du=\frac{1}{8}u^8+c=\frac{1}{8}ln^8(x)+c$

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Example Question #2600 : Calculus Ii

Evaluate the integrand:

$\int \frac{dx}{9x+7}dx$

Possible Answers:

$\frac{1}{9}ln(9x+7)+c$

$\frac{1}{9}ln{x}+7x+c$

ln(9x-7)+c

$ln(9x^{10}+7x)+c$

ln(9x+7)+c

Correct answer:

$\frac{1}{9}ln(9x+7)+c$

Explanation:

You have to use substitution to solve the integrand:

Let u=9x+7 and du=9dx and

$dx=\frac{du}{9}$

Therefore:

$\int \frac{dx}{9x+7}=\frac{1}{9}\int \frac{du}{u}=\frac{1}{9}ln(9x+7)+c$

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Example Question #2601 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(7sin(10x))}{4}+\frac{ (19x^{2})}{5})dx\end{align*}$

Possible Answers:

$\frac{(19x^{3})}{15}-\frac{ (7cos(10x))}{40}$

$\frac{(76x^{3})}{15}-\frac{ (35cos(10x))}{2}$

$\frac{(19x^{3})}{15}-\frac{ (7cos(10x))}{40}+C$

$\frac{(76x^{3})}{15}-\frac{ (35cos(10x))}{2}+C$

Correct answer:

$\frac{(19x^{3})}{15}-\frac{ (7cos(10x))}{40}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{The }a\text{ value for }\frac{(19x^{2})}{5}\text{ is }2\\&\text{The }a\text{ value for }\frac{(7sin(10x))}{4}\text{ is }10\\&\int(\frac{(7sin(10x))}{4}+\frac{ (19x^{2})}{5})dx=\frac{(19x^{3})}{15}-\frac{ (7cos(10x))}{40}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2602 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(3\cdot 3^{(\frac{x}{3})} + 11x^{15})dx\end{align*}$

Possible Answers:

$\frac{3^{(\frac{x}{3})}}{ln(3)}+\frac{ (2475x^{16})}{16}$

$\frac{(9\cdot 3^{(\frac{x}{3})})}{ln(3)}+\frac{ (11x^{16})}{16}+C$

$\frac{(9\cdot 3^{(\frac{x}{3})})}{ln(3)}+\frac{ (11x^{16})}{16}$

$\frac{3^{(\frac{x}{3})}}{ln(3)}+\frac{ (2475x^{16})}{16}+C$

Correct answer:

$\frac{(9\cdot 3^{(\frac{x}{3})})}{ln(3)}+\frac{ (11x^{16})}{16}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{The }a\text{ value for }11x^{15}\text{ is }15\\&\text{The }a\text{ value for }3\cdot 3^{(\frac{x}{3})}\text{ is }\frac{1}{3}\\&\int(3\cdot 3^{(\frac{x}{3})} + 11x^{15})dx=\frac{(9\cdot 3^{(\frac{x}{3})})}{ln(3)}+\frac{ (11x^{16})}{16}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2603 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(41sin(9x))}{10}+\frac{ (2\cdot 3^{(7x)})}{21})dx\end{align*}$

Possible Answers:

$\frac{(2\cdot 3^{(7x)})}{(3ln(3))}-\frac{ (369cos(9x))}{10}$

$\frac{(2\cdot 3^{(7x)})}{(147ln(3))}-\frac{ (41cos(9x))}{90}+C$

$\frac{(2\cdot 3^{(7x)})}{(3ln(3))}-\frac{ (369cos(9x))}{10}+C$

$\frac{(2\cdot 3^{(7x)})}{(147ln(3))}-\frac{ (41cos(9x))}{90}$

Correct answer:

$\frac{(2\cdot 3^{(7x)})}{(147ln(3))}-\frac{ (41cos(9x))}{90}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{The }a\text{ value for }\frac{(2\cdot 3^{(7x)})}{21}\text{ is }7\\&\text{The }a\text{ value for }\frac{(41sin(9x))}{10}\text{ is }9\\&\int(\frac{(41sin(9x))}{10}+\frac{ (2\cdot 3^{(7x)})}{21})dx=\frac{(2\cdot 3^{(7x)})}{(147ln(3))}-\frac{ (41cos(9x))}{90}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2604 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{7}{(1170x)}+\frac{ (45x^{7})}{8})dx\end{align*}$

Possible Answers:

$\frac{(7ln(x))}{1170}+\frac{ (45x^{8})}{64}$

$\frac{(187ln(x))}{1170}+\frac{ (45x^{8})}{8}$

$\frac{(187ln(x))}{1170}+\frac{ (45x^{8})}{8}+C$

$\frac{(7ln(x))}{1170}+\frac{ (45x^{8})}{64}+C$

Correct answer:

$\frac{(7ln(x))}{1170}+\frac{ (45x^{8})}{64}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{7}{(1170x)}\text{ is }\frac{7}{1170}\\&\text{The }a\text{ value for }\frac{(45x^{7})}{8}\text{ is }7\\&\int(\frac{7}{(1170x)}+\frac{ (45x^{7})}{8})dx=\frac{(7ln(x))}{1170}+\frac{ (45x^{8})}{64}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2605 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(3cos(x + 19))}{14}+\frac{ (38x^{6})}{3})dx\end{align*}$

Possible Answers:

$\frac{(1083sin(x + 19))}{14}+\frac{ (456x^{7})}{7}+C$

$\frac{(3sin(x + 19))}{14}+\frac{ (38x^{7})}{21}$

$\frac{(3sin(x + 19))}{14}+\frac{ (38x^{7})}{21}+C$

$\frac{(1083sin(x + 19))}{14}+\frac{ (456x^{7})}{7}$

Correct answer:

$\frac{(3sin(x + 19))}{14}+\frac{ (38x^{7})}{21}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{The }a\text{ value for }\frac{(38x^{6})}{3}\text{ is }6\\&\text{In deriving the inside of}\frac{(3cos(x + 19))}{14}\text{, the lone constant goes to zero.}\\&\int(\frac{(3cos(x + 19))}{14}+\frac{ (38x^{6})}{3})dx=\frac{(3sin(x + 19))}{14}+\frac{ (38x^{7})}{21}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2606 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{7}{(6x)}+\frac{ 4}{(9x^{6})})dx\end{align*}$

Possible Answers:

$\frac{(7ln(x))}{6}-\frac{ 4}{(45x^{5})}$

$\frac{(7ln(x))}{6}-\frac{ 4}{(45x^{5})}+C$

$\frac{(343ln(x))}{216}-\frac{ 16}{(5x^{5})}$

$\frac{(343ln(x))}{216}-\frac{ 16}{(5x^{5})}+C$

Correct answer:

$\frac{(7ln(x))}{6}-\frac{ 4}{(45x^{5})}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{7}{(6x)}\text{ is }\frac{7}{6}\\&\text{The }a\text{ value for }\frac{4}{(9x^{6})}\text{ is }-6\\&\int(\frac{7}{(6x)}+\frac{ 4}{(9x^{6})})dx=\frac{(7ln(x))}{6}-\frac{ 4}{(45x^{5})}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2607 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(13x^{6})}{3}+\frac{ (9x^{7})}{49})dx\end{align*}$

Possible Answers:

$\frac{(156x^{7})}{7}+\frac{ (9x^{8})}{8}+C$

$\frac{(156x^{7})}{7}+\frac{ (9x^{8})}{8}$

$\frac{(x^{7}\cdot (27x + 728))}{1176}+C$

$\frac{(x^{7}\cdot (27x + 728))}{1176}$

Correct answer:

$\frac{(x^{7}\cdot (27x + 728))}{1176}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{(9x^{7})}{49}\text{ is }7\\&\text{The }a\text{ value for }\frac{(13x^{6})}{3}\text{ is }6\\&\int(\frac{(13x^{6})}{3}+\frac{ (9x^{7})}{49})dx=\frac{(x^{7}\cdot (27x + 728))}{1176}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2598 : Calculus Ii

Example Question #2599 : Calculus Ii

Example Question #2600 : Calculus Ii

Example Question #2601 : Calculus Ii

Example Question #2602 : Calculus Ii

Example Question #2603 : Calculus Ii

Example Question #2604 : Calculus Ii

Example Question #2605 : Calculus Ii

Example Question #2606 : Calculus Ii

Example Question #2607 : Calculus Ii

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