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Calculus 2 : Indefinite Integrals

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Example Questions

Example Question #802 : Integrals

Evaluate the following indefinite integral:

$\int\sqrt{9-x^2}\, dx$

Possible Answers:

$\frac{\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}(x)+C$

$\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\cos^{-1}(\frac{x}{3})+C$

$\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}(\frac{x}{3})+C$

$\frac{\sqrt{9-x^2}}{2}+\frac{9}{2}\cos^{-1}(x)+C$

$\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\cos^{-1}(x)+C$

Correct answer:

$\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}(\frac{x}{3})+C$

Explanation:

Make the trigonometric substitution:

$x=3\sin(a),\: \: \:dx=3\cos(a)da$

Applying this to the expression given:

$\int\sqrt{9-x^2}\, dx=\int3\cos(a)\sqrt{9(1-\sin^2a)}\, da=\int9\cos^2(a)da$

$=9[\frac{1}{2}\cos(a)\sin(a)+\frac{a}{2}=\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}(\frac{x}{3})+C$

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Example Question #801 : Integrals

Evaluate the following indefinite integral:

$\int\frac{1}{x^2\sqrt{x^2-25}}\,dx$

Possible Answers:

$\frac{\sqrt{x^2-25}}{25x}+C$

$\frac{\sqrt{x^2-25}}{25}+C$

$\frac{\sqrt{x^2-25}}{5x}+C$

$\frac{\sqrt{x^2-25}}{5}+C$

$\frac{x^2-25}{5x}+C$

Correct answer:

$\frac{\sqrt{x^2-25}}{25x}+C$

Explanation:

Make the trigonometric substitution:

$x=5\sec(a),\,\,\,dx=5\tan(a)\sec(a)da$

Applying this substitution to the expression given, we can integrate:

$\int\frac{1}{x^2\sqrt{x^2-25}}\,dx=\int\frac{5\tan(a)\sec(a)}{25\sec^2(a) \sqrt{25\sec^2(a)-25}} da$

$=\int\frac{5\tan(a)}{125\sec(a)\tan(a)}da=\int\frac{da}{25\sec(a)}=\int\frac{\cos(a)}{25}da$

$=\frac{sin(a)}{25}+C=\frac{\sqrt{x^2-25}}{25x}+C$

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Example Question #802 : Integrals

Evaluate the following indefinite integral:

$\int \frac{2x-14}{x^2-2x-8}\,dx$

Possible Answers:

$3\ln(x+2)-\ln(x-4)+C$

$\frac{x^2-14x}{\frac{1}{3}x^3-x^2-8x}+C$

$-\ln(x+2)-3\ln(x-4)+C$

$5\ln(x+2)+3\ln(x-4)+C$

$3\ln(x+2)+3\ln(x-4)+C$

Correct answer:

$3\ln(x+2)-\ln(x-4)+C$

Explanation:

This integral can be solved by using partial fractions. First, we have to factor the denominator write the fraction as a sum of two fractions:

$\int \frac{2x-14}{x^2-2x-8}\,dx=\int \frac{2x-14}{(x+2)(x-4)}\,dx=\int(\frac{A}{x+2}+\frac{B}{x-4})dx$

Next, we can solve for A and B:

2x-14=A(x-4)+B(x+2)

When we let x=4:

$8-14=0+6B\rightarrow B=-1$

When x=-2:

$-4-14=-6A\rightarrow A=-1$

Replacing A and B in the integral, we can now solve it:

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

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Example Question #811 : 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 #2561 : Calculus Ii

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

Possible Answers:

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

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

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

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

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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Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #802 : Integrals

Example Question #801 : Integrals

Example Question #802 : Integrals

Example Question #811 : Integrals

Example Question #2561 : Calculus Ii

Example Question #182 : Indefinite Integrals

Example Question #2561 : Calculus Ii

Example Question #812 : Integrals

Example Question #813 : Integrals

Example Question #814 : Integrals

All Calculus 2 Resources

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