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Calculus 2 : Indefinite Integrals

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Example Questions

Example Question #738 : Integrals

$\int 2x^{\frac{1}{4}}dx$

Possible Answers:

$\frac{7}{5}x^{\frac{5}{4}}+C$

$\frac{8}{5}x^{\frac{3}{4}}+C$

$\frac{8}{5}x^{\frac{5}{4}}$

$\frac{8}{5}x^{\frac{5}{4}}+C$

Correct answer:

$\frac{8}{5}x^{\frac{5}{4}}+C$

Explanation:

To integrate this, first pull the 2 outside of the integral sign:

$2\int x^{\frac{1}{4}}dx$ .

Then, when integrating, remember to add 1 to the exponent and then also put that result on the denominator:

$2(\frac{x^{\frac{5}{4}}}{\frac{5}{4}})$ .

Then, simplify:

$2(\frac{4}{5})x^{\frac{5}{4}}=\frac{8}{5}x^\frac{5}{4}$ .

Then, add a +C because it is an indefinite integral:

$\frac{8}{5}x^{\frac{5}{4}}+C$ .

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Example Question #739 : Integrals

$\int \frac{x^3}{x^4-1}dx$

Possible Answers:

$\frac{1}{4}ln\left | x^4-1 \right |+C$

$ln\left | x^4-1 \right |+C$

$\frac{1}{4}ln\left | x \right |+C$

$\frac{1}{4}ln\left | x^4-1 \right |$

Correct answer:

$\frac{1}{4}ln\left | x^4-1 \right |+C$

Explanation:

Use u substitution here to integrate.

Assign

u=x^4-1 ,

$du=4x^3; \frac{1}{4}du=x^3$ .

Substitute in to rewrite the expression:

$\frac{1}{4}\int \frac{1}{u}du$ .

Recall that when integrating a single variable on the denominator, you take the natural log of that term.

Therefore, after integrating, you get

$\frac{1}{4}ln\left | u \right |$ .

Substitute in your initial expression to get

$\frac{1}{4}ln\left | x^4-1 \right |$ .

Remember to add a C because it is an indefinite integral:

$\frac{1}{4}ln\left | x^4-1 \right |+C$ .

Report an Error

Example Question #731 : Integrals

$\int 9x^{\frac{1}{3}}dx$

Possible Answers:

$\frac{27}{4}x^{\frac{3}{4}}+C$

$\frac{27}{4}x^{\frac{4}{3}}+C$

$9x^{\frac{4}{3}}+C$

$\frac{7}{4}x^{\frac{4}{3}}+C$

Correct answer:

$\frac{27}{4}x^{\frac{4}{3}}+C$

Explanation:

First, put the 9 on the outside of the integral sign:

$9\int x^{\frac{1}{3}}dx$ .

Then, when integrating, remember to raise the exponent by 1 and put that result on the denominator:

$9(\frac{x\frac{4}{3}}{\frac{4}{3}})$ .

Simplify:

$9(\frac{3}{4})x^{\frac{4}{3}}=\frac{27}{4}x^{\frac{4}{3}}$ .

Remember to add a plus C at the end because it is an indefinite integral:

$\frac{27}{4}x^{\frac{4}{3}}+C.$

Report an Error

Example Question #392 : Finding Integrals

Evaluate.

$\int \frac{1}{15x- 250} \ dx$

Possible Answers:

$F(x) =\frac{\ln(x-250)}{15} + C$

$F(x) =\ln(15x-250) + C$

Answer not listed.

$F(x) =\frac{x}{\ln(15x-250)} + C$

$F(x) =\frac{\ln(15x-250)}{15} + C$

Correct answer:

$F(x) =\frac{\ln(15x-250)}{15} + C$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{1}{15x- 250}$ .

The antiderivative is $F(x) =\frac{\ln(15x-250)}{15} + C$ .

Report an Error

Example Question #393 : Finding Integrals

Evaluate.

$\int \sin (3x - 7) \ dx$

Possible Answers:

$F(x) =-\cos(3x-7) + C$

Answer not listed.

$F(x) =\frac{\cos(3x+7)}{3} + C$

$F(x) =-\frac{x}{\cos(3x-7)} + C$

$F(x) =-\frac{\cos(3x-7)}{3} + C$

Correct answer:

$F(x) =-\frac{\cos(3x-7)}{3} + C$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \sin (3x - 7)$ .

The antiderivative is $F(x) =-\frac{\cos(3x-7)}{3} + C$ .

Report an Error

Example Question #394 : Finding Integrals

Evaluate.

$\int 6x +4 \ dx$

Possible Answers:

F(x) =3x + 4 + C

$F(x) =\frac{6x^3}{2} + 4x^2 + C$

Answer not listed.

F(x) =3x^2 + 4x + C

F(x) =6x^3 + 4x^2 + C

Correct answer:

F(x) =3x^2 + 4x + C

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, f(x) = 6x +4 .

The antiderivative is F(x) =3x^2 + 4x + C .

Report an Error

Example Question #114 : Indefinite Integrals

Evaluate:

$\int \frac{dx}{x^2-36}$

Possible Answers:

$\frac{1}{12}\ln\left | x+6\right |-\frac{1}{12}\ln\left | x-6\right |+C$

${12}\ln\left | x-6\right |-{12}\ln\left | x+6\right |+C$

$\frac{1}{12}\ln\left | x-6\right |-\frac{1}{12}\ln\left | x+6\right |$

$\frac{1}{12}\ln\left | x-6\right |+\frac{1}{12}\ln\left | x+6\right |+C$

$\frac{1}{12}\ln\left | x-6\right |-\frac{1}{12}\ln\left | x+6\right |+C$

Correct answer:

$\frac{1}{12}\ln\left | x-6\right |-\frac{1}{12}\ln\left | x+6\right |+C$

Explanation:

To integrate, we must use partial fraction decomposition (the degree of the denominator is larger than that of the numerator).

We will fully decompose the integrand, and then integrate:

$\frac{1}{x^2-36}=\frac{1}{(x-6)(x+6)}=\frac{A}{x-6}+\frac{B}{x+6}$

Note that the number of letters (2 - A and B) is the same as the degree of the polynomial we are decomposing. This will always hold true.

Now, solve for A and B:

1=A(x+6)+B(x-6)

Ax+6A+Bx-6B=1

Now, compare coefficients on either side of the equation:

A+B=0

6A-6B=1

$A=\frac{1}{12}, B=-\frac{1}{12}$

Now, rewrite the integrand as the sum of two fractions, and then splitting their sum into two separate integrals:

$\int \frac{dx}{12(x-6)}-\int \frac{dx}{12(x+6)}$

Finally integrate:

$\frac{1}{12}\ln\left | x-6\right |-\frac{1}{12}\ln\left | x+6\right |+C$

We used the following integration rule:

$\int \frac{dx}{x}=\ln\left | x\right |+C$

Report an Error

Example Question #2491 : Calculus Ii

Evaluate:

$\int \frac{dx}{x\sqrt{9x^2-16}}$

Possible Answers:

$\frac{1}{4}\arctan\frac{\left | 3x\right |}{4}+C$

$\frac{1}{4}\sec^{-1}\frac{\left | u\right |}{4}+C$

$\frac{1}{4}\sec^{-1}\frac{\left | 3x\right |}{4}+C$

$\frac{1}{3x}\sec^{-1}\frac{\left | 3x\right |}{4}+C$

$\frac{3}{4}\sec^{-1}\frac{\left | 3x\right |}{4}+C$

Correct answer:

$\frac{1}{4}\sec^{-1}\frac{\left | 3x\right |}{4}+C$

Explanation:

To evaluate the integral, we must first make a substitution:

u=3x

du=3dx

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Now, rewrite the integral in terms of u:

$\int \frac{du}{3x\sqrt{u^2-16}} = \int \frac{du}{u\sqrt{u^2-16}}$

Note that we have a 3x in the denominator which can be rewritten as u, as to make the integrand entirely in terms of u (and able to be solved).

Next, integrate:

$\int \frac{du}{u\sqrt{u^2-16}}=\frac{1}{4}\sec^{-1}\frac{\left | u\right |}{4}+C$

using the following rule:

$\int \frac{du}{u\sqrt{u^2=a^2}}=\frac{1}{a}\sec^{-1}\frac{\left | u\right |}{a}+C$

Finally, replace u with our original x term:

$\frac{1}{4}\sec^{-1}\frac{\left | 3x\right |}{4}+C$

Report an Error

Example Question #395 : Finding Integrals

Integrate:

$\int \frac{5x}{(x^2+4)(x+1)}dx$

Possible Answers:

$\frac{1}{2}\ln(x^2+4)+2\arctan(\frac{x}{2})-\ln(x+1)+C$

$\frac{1}{2}\ln(x^2+4)+\frac{1}{2}\arctan(\frac{x}{2})-\ln\left | x+1\right |+C$

$\ln(x^2+4)+2\arctan(\frac{x}{2})-\ln\left | x+1\right |+C$

$\frac{1}{2}\ln(x^2+4)+2\arctan(\frac{x}{2})-\ln\left | x+1\right |+C$

$\frac{1}{2}\ln(x^2+4)+2\arctan(\frac{x}{2})+\ln\left | x+1\right |+C$

Correct answer:

$\frac{1}{2}\ln(x^2+4)+2\arctan(\frac{x}{2})-\ln\left | x+1\right |+C$

Explanation:

To integrate, we must decompose the integrand into partial fractions:

$\frac{5x}{(x^2+4)(x+1)}=\frac{Ax+B}{x^2+4}+\frac{C}{x+1}$

Note that the total number of unknowns (3 - A, B, and C) is equal to the degree of the denominator before separating it (third degree polynomial). This will always be true!

Next, we solve for our unknowns by cross multiplying, and then comparing coefficients:

5x=(Ax+B)(x+1)+C(x^2+4)

5x=Ax^2+Ax+Bx+B+Cx^2+4C

The coefficients of the like terms on one side will sum to those on the other side of the equation:

A+C=0

A+B=5

B+4C=0

Solving, we get

A=1, B=4, C=-1

Putting these back into the fractions, and then rewriting the integrand as the sum of those fractions, we get

$\int \frac{x+4}{x^2+4}dx - \int \frac{dx}{x+1}=\int \frac{x}{x^2+4}dx+\int \frac{4}{x^2+4}dx-\int \frac{dx}{x+1}$

We split the fractions into smaller ones that can all be easily integrated.

The first integral can be solved using a substitution:

u=x^2+4

du=2xdx

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, rewrite the integral in terms of u and integrate:

$\frac{1}{2}\int \frac{du}{u}=\frac{1}{2}\ln\left | u\right |+C$

The integral was found using the following rule:

$\int \frac{dx}{x}=\ln\left | x\right |+C$

Now, replace u with the original x term:

$\frac{1}{2}\ln(x^2+4)+C$

Note that we removed the absolute value sign because what we have will always be positive.

The second integral is equal to

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

and was found using the following rule:

$\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan(\frac{x}{a})+C$

The third integral can be solved using a substitution:

u=x+1

du=dx

The derivative was found using the same rule as above.

Next, rewrite the integral in terms of u and integrate:

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

The integral was found using the same rule as above.

Finally, replace u with the original x term:

$\ln\left | x+1\right |+C$

Adding the integrals together, we get

$\frac{1}{2}\ln(x^2+4)+2\arctan(\frac{x}{2})-\ln\left | x+1\right |+C$

Report an Error

Example Question #396 : Finding Integrals

Evaluate.

$\int \frac{1}{3x-7} \ dx$

Possible Answers:

$F(x) =\frac{\ln(x-7)}{x} +C$

$F(x) =3\ln(3x-7) +C$

$F(x) =\frac{\ln(3x-7)}{3} +C$

$F(x) =\frac{x}{\ln(3x-7)} +C$

Answer not listed.

Correct answer:

$F(x) =\frac{\ln(3x-7)}{3} +C$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{1}{3x-7}$ .

The antiderivative is $F(x) =\frac{\ln(3x-7)}{3} +C$ .

Report an Error

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Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #738 : Integrals

Example Question #739 : Integrals

Example Question #731 : Integrals

Example Question #392 : Finding Integrals

Example Question #393 : Finding Integrals

Example Question #394 : Finding Integrals

Example Question #114 : Indefinite Integrals

Example Question #2491 : Calculus Ii

Example Question #395 : Finding Integrals

Example Question #396 : Finding Integrals

All Calculus 2 Resources

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