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Calculus 2 : Indefinite Integrals

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Example Questions

Example Question #21 : Indefinite Integrals

Solve: $\int\:3x^2+ln(x) \: dx$

Possible Answers:

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

x^3-xln(x)-x+C

x^3+xln(x)+x+C

x^3+xln(x)-x+C

$x^3+\frac{1}{x}+C$

Correct answer:

x^3+xln(x)-x+C

Explanation:

The indefinite integral can be split into two separate integrals.

$\int\:3x^2+ln(x) \: dx = \int\:3x^2\:dx+\int ln(x) \: dx$

Evaluate each integral.

$\int\:3x^2\:dx = \frac{3x^{2+1}}{2+1} +C_1 = x^3+C_1$

The answer to the first integral is: x^3+C_1

Be careful since the answer to $\int ln(x) \: dx$ is not $\frac{1}{x}$ as this is the derivative of ln(x) . In order to solve this integral, we will need to use integration by parts.

If we let u=ln(x) , we will have $du= \frac{1}{x} dx$ by differentiation, and if we let dv=dx , we will have v=x by integration. The constant can be added in at the end of the problem.

Write the formula for integration by parts.

$\int u\:dv = uv-\int v\:du$

Substitute the terms into the formula.

$\int u\:dv =ln(x)\cdot x-\int(x\cdot \frac{1}{x}\:dx)$

Simplify the terms inside the integral and evaluate.

$\int u\:dv =xln(x)-\int\:dx$

$\int u\:dv =xln(x)-x+C_2$

The answer to the second integral is: xln(x)-x+C_2

Combine the two answers. The constants C_1, C_2 can be combined to be a single constant term at the end.

The answer is: x^3+xln(x)-x+C

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Example Question #21 : Indefinite Integrals

Evaluate: $\int 2x\sqrt{x-1}\:dx$

Possible Answers:

$\frac{21}{15}(x-1)^{\frac{4}{3}}+\frac{16}{3}(x-1)^{\frac{3}{2}} +C$

$\frac{2}{5}(x-1)^{\frac{1}{3}}+\frac{3}{4}(x-1)^{\frac{5}{2}} +C$

$\frac{4}{5}(x-1)^{\frac{5}{2}}-\frac{4}{3}(x-1)^{\frac{3}{2}} +C$

$\frac{4}{3}x(x-1)^{\frac{3}{2}}+C$

$\frac{4}{5}(x-1)^{\frac{5}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}} +C$

Correct answer:

$\frac{4}{5}(x-1)^{\frac{5}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}} +C$

Explanation:

This integral will require substitution with both and terms.

If we let u=x-1 , then x =u+1 .

Differentiate with respect to .

du=dx

Substitute all terms back into the integral.

$\int 2x\sqrt{x-1}\:dx = \int 2(u+1)\sqrt{u}\:du$

The $\sqrt u$ term is also the same as $u^{\frac{1}{2}}$ , which can be multiplied within the terms of the parentheses. Simplify the integral.

$\int 2(u+1)\sqrt{u}\:du = \int 2u^{\frac{3}{2}}+2u^{\frac{1}{2}}\: du$

Evaluate this integral.

$\int 2u^{\frac{3}{2}}+2u^{\frac{1}{2}}\: du = 2[u^{\frac{5}{2}}\cdot \frac{2}{5}] + 2[u^{\frac{3}{2}}\cdot \frac{2}{3}] =\frac{4}{5}u^{\frac{5}{2}}+\frac{4}{3}u^{\frac{3}{2}}+C$

Resubstitute .

$\frac{4}{5}u^{\frac{5}{2}}+\frac{4}{3}u^{\frac{3}{2}} +C= \frac{4}{5}(x-1)^{\frac{5}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}} +C$

The answer is: $\frac{4}{5}(x-1)^{\frac{5}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}} +C$

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Example Question #23 : Indefinite Integrals

Evaluate: $\int \frac{6}{x^2+4x+10}\:dx$

Possible Answers:

$\frac{\sqrt6}{6} tan^{-1}(\frac{x+2}{\sqrt6})+C$

$\sqrt6\:sec^{-1}(\frac{x+2}{\sqrt6})+C$

$\sqrt6\:tan^{-1}(\frac{x+2}{\sqrt6})+C$

$\frac{\sqrt{6}\: ln\left | x^2+4x+10\right |}{2x+4}+C$

$\frac{6ln\left | x^2+4x+10\right |}{2x+4}+C$

Correct answer:

$\sqrt6\:tan^{-1}(\frac{x+2}{\sqrt6})+C$

Explanation:

The denominator is irreducible, which means that we cannot use partial fractions to identify the coefficients of the separable fractions. The only method we can use is to complete the square and use the identity of inverse trigonometry.

Complete the square for the denominator. This is done by taking the square of half of the middle term, and then subtracting this digit on the end.

x^2+4x+10 = (x^2+4x+4)+10 -(4)

Factorize the parabolic function in the parentheses and simplify.

x^2+4x+10 = (x+2)^2+6

Rewrite the integral and pull out the six in front of the integral.

$\int \frac{6}{x^2+4x+10}\:dx = 6\int \frac{1}{(x+2)^2+6}\:dx$

Write the integral property of inverse tangent.

$\frac{1}{A} tan^{-1}(\frac{X}{A})+C = \int \frac{1}{X^2+A^2}$

By this rule, substitute the terms and simplify to get the and terms.

X^2 = (x+2)^2

X=x+2

A^2 = 6

$A=\sqrt6$

Substitute the and terms into the inverse tangent rule.

$\frac{1}{\sqrt6} tan^{-1}(\frac{x+2}{\sqrt6})+C$

Rationalize the denominator of the coefficient.

$\frac{1}{\sqrt6} \cdot \frac{\sqrt6}{\sqrt6} tan^{-1}(\frac{x+2}{\sqrt6})+C = \frac{\sqrt6}{6} tan^{-1}(\frac{x+2}{\sqrt6})+C$

Substituting this back in the original integral, this means that:

$6\int \frac{1}{(x+2)^2+6}\:dx = 6(\frac{\sqrt6}{6} tan^{-1}(\frac{x+2}{\sqrt6})+C)$

Do not forget to multiply the six that is outside the integral.

$= 6(\frac{\sqrt6}{6} tan^{-1}(\frac{x+2}{\sqrt6})+C) = \sqrt6\:tan^{-1}(\frac{x+2}{\sqrt6})+C$

The answer is: $\sqrt6\:tan^{-1}(\frac{x+2}{\sqrt6})+C$

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Example Question #24 : Indefinite Integrals

Find the integral of

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

Possible Answers:

$\frac{2}{x} + C$

$ln|x| - \frac{1}{x} + C$

$-ln|x| + \frac{2}{x} + C$

$ln|x| + \frac{2}{x} + C$

$ln|x| - \frac{2}{x} + C$

Correct answer:

$ln|x| - \frac{2}{x} + C$

Explanation:

Simplify:

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

Integrate:

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

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Example Question #21 : Indefinite Integrals

$\int ln(x)dx$

Possible Answers:

$\frac{1}{x^2}$

$\frac{1}{x}$

x(ln(x)-1)+C

$ln(x)+\frac{1}{x}$

x(ln(x)+1)+C

Correct answer:

x(ln(x)-1)+C

Explanation:

Use Integration by parts:

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

dv = dx ; v = x

$\int ln(x)dx = \int udv = uv-\int vdu$

= $xln(x) - \int x(\frac{1}{x}dx) = xln(x) - x + C$

= x(ln(x)-1)+C

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Example Question #26 : Indefinite Integrals

Find the indefinite integral

$\int {4}x^3\, dx$

Possible Answers:

x^4 +c

$\frac{x^4}{4}+c$

12x^2 +c

DNE

Correct answer:

x^4 +c

Explanation:

To find the indefinite integral, we use the inverse power rule which states

$\int x^n \, dx = \frac{x^{n+1}}{n+1}$

For the problem in this question,

$\int \4x^3\, dx = {4} \int x^3 \, dx$

$=4 (\frac{x^4}{4}) +c$

=x^4 + c

As such,

$\int {4}x^3\, dx = x^4 +c$

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Example Question #27 : Indefinite Integrals

Find the indefinite integral

$\int [sec(x)tan(x)+sec^2(x)]\, dx$

Possible Answers:

tan(x)+c

sec(x)tan(x)+c

sec(x)+1+c

sec(x)+tan(x)+c

Correct answer:

sec(x)+tan(x)+c

Explanation:

Because integration is a linear operation, we are able to anti-differentiate the function term by term.

We use the properties that

The anti-derivative of is
The anti-derivative of is

to solve the indefinite integral

$\int [sec(x)tan(x)+sec^2(x)]\, dx = sec(x)+tan(x)+c$

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Example Question #28 : Indefinite Integrals

Evaluate the following integral:

$\int 2xe^xdx$

Possible Answers:

2xe^x-2e^x+C

2xe^x+2e^x+C

2x+e^x+C

2xe^x+4e^x+C

Correct answer:

2xe^x-2e^x+C

Explanation:

To integrate, we must integrate by parts, according to the formula

$\int udv=uv-\int vdu$

Now, we choose our u (from which we get du), and dv (from which we get v):

u=2x, du=2dx

dv=e^xdx, v=e^x

The rules for the derivation and integration are:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\int e^x dx=e^x+C$

(Note that we do not include the constant of integration.)

Use the above formula, and integrate:

$2xe^x-\int 2e^xdx=2xe^x-2e^x+C$

The integral was performed using the same rule as stated above.

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Example Question #29 : Indefinite Integrals

Evaluate the indefinite integral $\int \frac{1}{x(x^2-1)}dx$ .

Possible Answers:

None of the other answers

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

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

$\frac{2}{3}\ln(x)-\frac{1}{2}\ln(x-1)+C$

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

Correct answer:

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

Explanation:

This integral can be evaluated using partial fraction decomposition as follows.

$\int \frac{1}{x(x^2-1)}dx$ . Start

$\int \frac{1}{x(x-1)(x+1)}dx$ . Factor the denominator completely.

Now use the method of partial fraction decomposition

$\frac{1}{x(x-1)(x+1)} = \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$

Multiply both sides by x(x-1)(x+1) , and simplify.

1= A(x^2-1)+B(x^2-x)+C(x^2+x)

Distribute A, B, C .

1=Ax^2-A+Bx^2-Bx+Cx^2+Cx

By equating like coefficents, we can rewite the above as a system of equations

-A =1

-B+C=0

A+B+C=0

Using any method you'd like to solve this system of equations, we obtain $A=1, B=\frac{1}{2}, C=\frac{1}{2}$ .

Substituting this back into our original integral, we obtain

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

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

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Example Question #30 : Indefinite Integrals

Determine the following integral:

$\int (3tan(2x)sec(2x))$

Possible Answers:

$\frac{3}{2}sec(2x)$

$\frac{3}{2}sec(x)$

$\frac{3}{2}csc(2x)$

3sec(2x)

Correct answer:

$\frac{3}{2}sec(2x)$

Explanation:

To determine this indefinite integal, we integrate by substitution:

We can substitute for .

u=2x

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}2x=2$

We can rewrite the original integral as:

$\int 3tan(2x)sec(2x) dx=\int 3tan(u)sec(u)du=$

$=\int\frac{3}{2}tan(u)sec(u)du$

Recall that $\int tan(x)sec(x) dx= sec(x)+C$ , where is a constant

Therefore:

$\int\frac{3}{2}tan(u)sec(u)du = \frac{3}{2}sec(u) +C$

Since

u=2x

${3}sec(u) du +C= \frac{3}2{}sec(2x)+C$ , where is a constant

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Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : Indefinite Integrals

Example Question #21 : Indefinite Integrals

Example Question #23 : Indefinite Integrals

Example Question #24 : Indefinite Integrals

Example Question #21 : Indefinite Integrals

Example Question #26 : Indefinite Integrals

Example Question #27 : Indefinite Integrals

Example Question #28 : Indefinite Integrals

Example Question #29 : Indefinite Integrals

Example Question #30 : Indefinite Integrals

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