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Calculus 2 : Integrals

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Example Questions

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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Example Question #2401 : Calculus Ii

$\int 3x^2+2x+4 \;dx$

Possible Answers:

$\frac{x^3}{3}+x^2+C$

x^3+x^2+C

x^3+x^2+4x+C

6x+2+C

6x+2

Correct answer:

x^3+x^2+4x+C

Explanation:

When integrating, we do the opposite of a derivative. You increase the exponent by one and divide the function by that new power. Since this is an indefinite integral, we have to add a at the end of the equation.

$\int 3x^2+2x^2+4dx=x^3+x^2+4x+C.$

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Example Question #2402 : Calculus Ii

$\int 5sin(x)-2cos(x)\;dx$

Possible Answers:

5cos(x)-2sin(x)+C

5cos(x)+2sin(x)+C

-5cos(x)-2sin(x)+C

-5cos(x)+2sin(x)+C

Correct answer:

-5cos(x)-2sin(x)+C

Explanation:

The antiderivative of sin(x)=-cos(x) . The antiderivative of cos(x)=sin(x) . Remember, this is the opposite of a derivative. Therefore, for our integral, we have:

$\int 5sin(x)-2cos(x)\;dx=-5cos(x)-2sin(x)+C$ .

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Example Question #651 : Integrals

Possible Answers:

4x^4+sin(x)+2x^2+C

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

$\frac{x^4}{4}+cos(x)+\frac{x^2}{2}$

x^4+sin(x)+x^2+C

x^4+cos(x)+x^2+C

Correct answer:

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

Explanation:

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

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

Example Question #2401 : Calculus Ii

Example Question #2402 : Calculus Ii

Example Question #651 : Integrals

All Calculus 2 Resources

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