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Calculus 2 : Finding Integrals

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Example Questions

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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Example Question #301 : Finding Integrals

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

Possible Answers:

6x+2

x^3+x^2+C

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

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

6x+2+C

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 #302 : Finding Integrals

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

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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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

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

Example Question #301 : Finding Integrals

Example Question #302 : Finding Integrals

All Calculus 2 Resources

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