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Calculus 2 : Indefinite Integrals

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Example Question #1 : Indefinite Integrals

Find the indefinite integral of the following function:

$f(x)=\frac{2x}{x^2+1}dx$

Possible Answers:

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

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

ln(x^2+1)

$\frac{2(x^2-1)}{(x^2+1)^2}$

ln|x^2+1|+C

Correct answer:

ln|x^2+1|+C

Explanation:

To integrate this function, use u substitution. Make,

$\\u=x^2+1 \\du=2xdx$

then substitute them into the equation to get

$f(x)=\frac{du}{u}$ .

The integral of

$\frac{du}{u}=ln|u|+C$

then plug u back into the equation

ln|x^2+1|+C .

The +C is essential because the integral is indefinite.

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Example Question #1 : Indefinite Integrals

Evaluate the given indefinite integral

$\int sec^2(x)tan(x)dx$ .

Possible Answers:

$\frac{sec^2(x)}{2}+C$

$\frac{cos^2(x)}{2}+C$

-ln(tan(x))+C

$\frac{sec^2(x)}{2}+C$

$\frac{sec^2(x)}{2}$

Correct answer:

$\frac{sec^2(x)}{2}+C$

Explanation:

To integrate this function, use u substitution. Make

$\\u=sec(x) \\du=sec(x)tan(x)dx$

then substitute them into the equation to get

$\int udu$ .

The integral of

$udu=\frac{u^2}{2}+C$

then plug u back into the equation

$\frac{sec^2(x)}{2}+C$ .

The +C is essential because the integral is indefinite.

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Example Question #3 : Indefinite Integrals

Evaluate the given indefinite integral

$\int -\frac{x}{\sqrt{1-x^2}}$ .

Possible Answers:

$\sqrt{1-x^2}+C$

$\sqrt{1-x^2}$

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

$\frac{1}{\sqrt{1-x^2}}+C$

${1-x^2}$

Correct answer:

$\sqrt{1-x^2}+C$

Explanation:

To integrate this function, use u substitution. Make

$\\u=1-x^2 \\du=-2xdx$

then substitute them into the equation to get

$\int \frac{\frac{1}{2}du}{\sqrt{u}}$ .

The integral of

$\int \frac{du}{\sqrt{u}}=2\sqrt{u}+C$

so we have

$\frac{1}{2}\cdot2\sqrt{u}+C=\sqrt{u}+C=\sqrt{1-x^2}+C$ .

The +C is essential because the integral is indefinite.

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Example Question #1 : Indefinite Integrals

Calculate the following indefinite integral.

$\int xe^{3x}dx$

Possible Answers:

$\frac{1}{3}x(e^{3x}+\frac{1}{3}e^{3x})+c$

$\frac{1}{3}xe^{9x} +\frac{1}{9}e^{3x}+c$

$\frac{1}{3}e^{3x} +\frac{1}{9}e^{3x}+c$

$\frac{1}{3}xe^{3x} +\frac{1}{9}e^{3x}+c$

$\frac{1}{9}e^{3x}(x+1)+c$

Correct answer:

$\frac{1}{3}xe^{3x} +\frac{1}{9}e^{3x}+c$

Explanation:

To calculate the integral, we need to use integration by parts. The definition for integration by parts is

$\int udv=uv-\int vdu$

It is important here to select the correct u and dv terms from our orginal integral. We eventually want of the terms to "go away" when we take its derivate. We notice here that out of our two functions in our integral, and $e^{3x}$ , the derivate of x is 1, making is very simple to integrate eventually. Therefore, will be our term, and $e^{3x}dx$ will be our dv term. Note that the dv term is not just dx, but the function attached to it as well. If $e^{3x}$ was our term, then xdx would be our dv term.

u=x $dv=e^{3x}dx$

Now calculate the terms and needed to proceed with the integration by parts equation.

du=dx $v=\int e^{3x}dx=\frac{1}{3}e^{3x}$

(you can to set integration constant c=0)

Now that we have the terms that we need, we can plug in these terms into the integration by parts formula above.

$(x)\left(\frac{1}{3}e^{3x}\right)$ - $\int \frac{1}{3}e^{3x}dx$

Note that although we still need to integrate one more time, this new integral only consists of one function which is simple to integrate, as opposed to the two functions we had before. Also note that the x term from the initial integral "went away", thus making the resulting integral easy to calculate.

Simplifying this term now becomes

$\frac{1}{3}xe^{3x}+\frac{1}{9}e^{3x}+c$ .

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Example Question #5 : Indefinite Integrals

$g(x)= \sqrt{2(1-cos^2(\sqrt2x))}$

What is

$\int g(x)dx$ ?

Possible Answers:

$-cos(\sqrt2x)+c$

$cos(\sqrt2x)+c$

-2cos(2x)+C

$\sqrt2sin(\sqrt2x)+c$

$\sqrt2cos(\sqrt2x)+c$

Correct answer:

$-cos(\sqrt2x)+c$

Explanation:

Although the integral looks difficult, it can be majorly simplified. Remember this crucial trig identity.

1-cos^2(u)=sin^2(u)

Using this identity, the integral can now be simplified to

$\int\sqrt{2sin^2(\sqrt2x)}dx=\sqrt2 \int sin(x)dx$ , which is very simple to integrate.

$=\sqrt2 \int sin(\sqrt2 x)dx$

$=-cos(\sqrt2x)+c$

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Example Question #6 : Indefinite Integrals

Solve the following for .

$\frac{dy}{e^x}=\frac{dx}{y}$

Assume the integration constant is zero.

Possible Answers:

$y=\sqrt{e^x}$

y=2e^x

$y=\pm\sqrt{2e^x}$

$y=\pm\frac{\sqrt{2e^x}}{x}$

$y=\pm\sqrt{e^2x}$

Correct answer:

$y=\pm\sqrt{2e^x}$

Explanation:

In this problem we can try to get all the terms with on one side and all the terms with on the other.

ydy=e^xdx

Now we can integrate both sides using the definition of and the power rule.

$\int x^n=\frac{x^{n+1}}{n+1}$ and $\int e^x=e^x$

$\int ydy=\int e^xdx$

$\frac{y^2}{2}=e^x +C,C=0$

y^2=2e^x

$y=\pm\sqrt{2e^x}$

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Example Question #7 : Indefinite Integrals

Solve the following for .

y(x+2)dy=dx

Assume the integration constant C=0.

Possible Answers:

y=ln(x)

$y=\pm\sqrt{ln(x+2)}$

$y=\pm\sqrt{2ln(x+2)}$

$y=\pm\2ln(x+2)$

$y=\sqrt{2ln(x)}$

Correct answer:

$y=\pm\sqrt{2ln(x+2)}$

Explanation:

Move all expressions with to one side and all to the other.

$y{dy}=\frac{dx}{x+2}$

Now integrate both sides using the power rule and the definition of natural log.

The power rule for integrals states,

$\int x^n =\frac{x^{n+1}}{n+1}$ and the definition of natural log is,

$\int \frac{1}{x}=ln(x)$ .

Applying these rules we are able to solve the problem.

$\frac{y^2}{2}=ln(x+2)$

$y=\pm\sqrt{2ln(x+2)}$

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Example Question #8 : Indefinite Integrals

Evaluate the following integral.

$\int sin^2({2x})dx$

Possible Answers:

$\frac{x}{2}-\frac{sin(4x)}{4}+C$

$\frac{x}{2}-\frac{sin(4x)}{8}+C$

${x}-\frac{sin(4x)}{4}+C$

$\frac{x}{4}-\frac{cos(4x)}{8}+C$

$\frac{x}{4}-\frac{cos(x)}{8}+C$

Correct answer:

$\frac{x}{2}-\frac{sin(4x)}{8}+C$

Explanation:

We need to use the following identity:

$sin^2(x)=\frac{1-cos(2x)}{2}$

Our integral now becomes

$\frac{1}{2} \int \1-cos(4x)dx$ .

Notice inside the cosine becomes 4 because we already had a 2 in the original expression.

This can be split into two integrals

$\frac{1}{2} \int 1dx +\frac{1}{2}\int \cos(4x)dx$ .

Which becomes

$\frac{1}{2}\left(x-\frac{sin({4x)}}4\right)+C$ .

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Example Question #9 : Indefinite Integrals

Find the indefinite integral of $f'(x)=9x^{2}+6x-12$ .

Possible Answers:

$f(x)=3x^{3}+3x^{2}-12x$

f(x)=18x+6

f(x)=18x+6+C

None of the above

$f(x)=3x^{3}+3x^{2}-12x+C$

Correct answer:

$f(x)=3x^{3}+3x^{2}-12x+C$

Explanation:

We can find the indefinite integral of $f'(x)=9x^{2}+6x-12$ using the Power Rule for Integrals, which states that

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for all $n\neq-1$ and with the arbitrary constant of integration .

Applying this rule to

$\int f'(x)dx=\int (9x^{2}+6x-12)dx=3x^{3}+3x^{2}-12x+C$ .

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Example Question #10 : Indefinite Integrals

Find the indefinite integral of $f'(x)=12x^{2}-2x+7$ .

Possible Answers:

f(x)=24x-2

None of the above

f(x)=24x-2+C

$f(x)=4x^{3}-x^{2}+7x+C$

$f(x)=4x^{3}-x^{2}+7x$

Correct answer:

$f(x)=4x^{3}-x^{2}+7x+C$

Explanation:

We can find the indefinite integral of $f'(x)=12x^{2}-2x+7$ using the Power Rule for Integrals, which states that

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for all $n\neq-1$ and with the arbitrary constant of integration .

Applying this rule to

$\int f'(x)dx=\int (12x^{2}-2x+7)dx=4x^{3}-x^{2}+7x+C$ .

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Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Indefinite Integrals

Example Question #1 : Indefinite Integrals

Example Question #3 : Indefinite Integrals

Example Question #1 : Indefinite Integrals

Example Question #5 : Indefinite Integrals

Example Question #6 : Indefinite Integrals

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