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Calculus 2 : Integrals

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Example Questions

Example Question #184 : Indefinite Integrals

Solve the following integral: $\int \frac{5dx}{(x+1)(x+2)}$

Possible Answers:

$5ln\left |\frac{x-1}{x+2} \right |+C$

$5ln\left |\frac{x+1}{x-2} \right |+C$

$ln\left |\frac{x+1}{x+2} \right |+C$

$5ln\left |\frac{x+1}{x+2} \right |+C$

Correct answer:

$5ln\left |\frac{x+1}{x+2} \right |+C$

Explanation:

$\int \frac{5dx}{(x+1)(x+2)}$

Solve using partial fraction decomposition

Factor out the 5 from the numerator: $5\int \frac{dx}{(x+1)(x+2)}$

Factor the integrand into its component partial fractions: $\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}$

Multiply by the denominators to eliminate the fractions: 1=A(x+2)+B(x+1)

Plug x=-2 into the equation to eliminate the A term: 1=0+B(-1)

Solve for B: B=-1

Plug x=-1 into 1=A(x+2)+B(x+1) , to eliminate B term: 1=A(1)+0 .

Solve for A: A=1 .

Re-write integrand with values for A and B: $\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2}$ .

Calculate $5\int \frac{dx}{(x+1)(x+2)}$ with integrand in its partial fraction decomposition: $5\int(\frac{1}{x+1}-\frac{1}{x+2})dx=5(ln\left |x+1 \right |-ln\left | x+2\right |)+C$ .

Simplify: $5(ln\left |x+1 \right |-ln\left | x+2\right |)+C=5ln\left | \frac{x+1}{x+2} \right |+C$

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Example Question #2561 : Calculus Ii

Calculate the following integral: $\int \frac{x^{3}+8}{x^2+3x+2}dx$

Possible Answers:

$x^{2}-3x+8ln\left | x+2\right |-ln\left | x+1\right |+C$

$x^{2}-x+8ln\left | x+2\right |-ln\left | x+1\right |+C$

$\frac{1}{2}x^{2}-x+8ln\left | x+2\right |-ln\left | x+1\right |+C$

$\frac{1}{2}x^{2}-3x+7ln\left | x+1\right |+C$

Correct answer:

$\frac{1}{2}x^{2}-3x+7ln\left | x+1\right |+C$

Explanation:

$\int \frac{x^{3}+8}{x^2+3x+2}dx$

Factor the numerator and denominator of the integrand:

$\int \frac{x^{3}+8}{x^2+3x+2}dx=\int \frac{(x+2)(x^{2}-2x+4)}{(x+2)(x+1)}dx$

Simplify the integrand: $\int \frac{(x+2)(x^{2}-2x+4)}{(x+2)(x+1)}dx=\int \frac{x^{2}-2x+4}{x+1}dx$

Use polynomial long division to make the degree of the numerator of the integrand less than the denominator: $\int \frac{x^{2}-2x+4}{x+1}dx=\int (x-3+ \frac{7}{x+1})dx$

Apply the difference rule to the integrand: $\int (x-3+ \frac{7}{x+1})dx=\int xdx-\int 3dx+\int \frac{7}{x+1}dx$

Solve the first 2 integrals: $\int xdx-\int 3dx=\frac{1}{2}x^{2}-3x+C$

Make the following substitution to solve the third integral: u=x+1 du=dx

Apply the substitution to the integral: $\int \frac{7}{x+1}dx=7\int \frac{du}{u}$

Solve the integral: $7\int \frac{du}{u}=7ln\left | u\right |+C$

Plug the value for u back into the expression: $7ln\left | u\right |+C=7ln\left | x+1\right +C|$

Combine the answers from the other two integrals:

$\int xdx-\int 3dx+\int \frac{7}{x+1}dx=\frac{1}{2}x^{2}-3x+7ln\left | x+1\right |+C$

Solution: $\int \frac{x^{3}+8}{x^2+3x+2}dx=\frac{1}{2}x^{2}-3x+7ln\left | x+1\right |+C$

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Example Question #182 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(9e^{(-12x)})}{32})dx\end{align*}$

Possible Answers:

$-\frac{(9e^{(-12x)})}{32}$

$-\frac{(27e^{(-12x)})}{8}$

$-\frac{(3e^{(-12x)})}{128}+C$

$-\frac{(27e^{(-12x)})}{8}+C$

Correct answer:

$-\frac{(3e^{(-12x)})}{128}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\text{Note for this problem, the }a \text{ value is} -12\\&\int(\frac{(9e^{(-12x)})}{32})dx=-\frac{(3e^{(-12x)})}{128}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #811 : Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(4sin(x + 6))}{29})dx\end{align*}$

Possible Answers:

$-\frac{(144cos(x + 6))}{29}$

$-\frac{(144cos(x + 6))}{29}+C$

$-\frac{(4cos(x + 6))}{29}+C$

$-\frac{(4cos(x + 6))}{29}$

Correct answer:

$-\frac{(4cos(x + 6))}{29}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{Note in deriving the inside, the lone constant goes to zero.}\\&\int(\frac{(4sin(x + 6))}{29})dx=-\frac{(4cos(x + 6))}{29}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #812 : Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(12sin(9x))}{7})dx\end{align*}$

Possible Answers:

$-\frac{(12cos(9x))}{7}$

$-\frac{(4cos(9x))}{21}+C$

$-\frac{(108cos(9x))}{7}$

$-\frac{(108cos(9x))}{7}+C$

Correct answer:

$-\frac{(4cos(9x))}{21}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{Note for this problem, the }a\text{ value is } 9\\&\int(\frac{(12sin(9x))}{7})dx=-\frac{(4cos(9x))}{21}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #813 : Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(9e^{(13x)})}{25})dx\end{align*}$

Possible Answers:

$\frac{(117e^{(13x)})}{25}+C$

$\frac{(9e^{(13x)})}{325}$

$\frac{(9e^{(13x)})}{325}+C$

$\frac{(117e^{(13x)})}{25}$

Correct answer:

$\frac{(9e^{(13x)})}{325}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\text{Note for this problem, the }a\text{ value is } 13\\&\int(\frac{(9e^{(13x)})}{25})dx=\frac{(9e^{(13x)})}{325}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #814 : Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(11e^{(2x)})}{21})dx\end{align*}$

Possible Answers:

$\frac{(11e^{(2x)})}{42}+C$

$\frac{(11e^{(2x)})}{42}$

$\frac{(22e^{(2x)})}{21}+C$

$\frac{(22e^{(2x)})}{21}$

Correct answer:

$\frac{(11e^{(2x)})}{42}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\text{Note for this problem, the }a\text{ value is } 2\\&\int(\frac{(11e^{(2x)})}{21})dx=\frac{(11e^{(2x)})}{42}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #191 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(5sin(x + 12))}{43})dx\end{align*}$

Possible Answers:

$-\frac{(5cos(x + 12))}{43}$

$-\frac{(5cos(x + 12))}{43}+C$

$-\frac{(720cos(x + 12))}{43}+C$

$-\frac{(720cos(x + 12))}{43}$

Correct answer:

$-\frac{(5cos(x + 12))}{43}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{Note in deriving the inside, the lone constant goes to zero.}\\&\int(\frac{(5sin(x + 12))}{43})dx=-\frac{(5cos(x + 12))}{43}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #192 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(2cos(16x))dx\end{align*}$

Possible Answers:

$\frac{sin(16x)}{8}$

32sin(16x)

32sin(16x)+C

$\frac{sin(16x)}{8}+C$

Correct answer:

$\frac{sin(16x)}{8}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{Note for this problem, the }a\text{ value is } 16\\&\int(2cos(16x))dx=\frac{sin(16x)}{8}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #193 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(9sin(x + 8))}{26})dx\end{align*}$

Possible Answers:

$-\frac{(288cos(x + 8))}{13}+C$

$-\frac{(9cos(x + 8))}{26}+C$

$-\frac{(288cos(x + 8))}{13}$

$-\frac{(9cos(x + 8))}{26}$

Correct answer:

$-\frac{(9cos(x + 8))}{26}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{Note in deriving the inside, the lone constant goes to zero.}\\&\int(\frac{(9sin(x + 8))}{26})dx=-\frac{(9cos(x + 8))}{26}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #184 : Indefinite Integrals

Example Question #2561 : Calculus Ii

Example Question #182 : Indefinite Integrals

Example Question #811 : Integrals

Example Question #812 : Integrals

Example Question #813 : Integrals

Example Question #814 : Integrals

Example Question #191 : Indefinite Integrals

Example Question #192 : Indefinite Integrals

Example Question #193 : Indefinite Integrals

All Calculus 2 Resources

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