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Calculus 2 : Finding Integrals

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Example Questions

Example Question #431 : Finding Integrals

Evaluate the integral:

$\int \frac{x^3-4x^2+2x+1}{x-2} dx$

Possible Answers:

$\frac{1}{3}x^3-x^2-2x+5\ln\left | x-2\right |+C$

$\frac{1}{3}x^3-x^2-2x-3\ln\left | x-2\right |$

$\frac{1}{3}x^3-x^2-2x+3\ln\left | x-2\right |+C$

$\frac{1}{3}x^3-x^2-2x-3\ln\left | x-2\right |+C$

Correct answer:

$\frac{1}{3}x^3-x^2-2x-3\ln\left | x-2\right |+C$

Explanation:

No substitution would work here, but rewriting the integrand via long division is easiest way to solve this problem.

After long division, the quotient should be x^2-2x-2 with a remainder of 3. Whenever there is a remainder, it must be placed over the divisor as a numerator. This makes the original integral expressed as:

$\int (x^2-2x-2-\frac{3}{x-2})dx$

Each term of the integrand can be integrated by using the following formula:

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

For the last term, however, a u-substitution is needed where u=x-2, du=dx . The integrand can be rewritten as:

$-3\int\frac{1}{u} du=-3\ln\left | u\right |+C=-3\ln\left | x-2\right |+C$

The final answer is:

$\frac{1}{3}x^3-x^2-2x-3\ln\left | x-2\right |+C$

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Example Question #432 : Finding Integrals

$\int \frac{10x^2-8x+16}{2x}dx$

Possible Answers:

$\frac{5}{2}x^2-4x+8ln\left | x \right |+C$

$\frac{5}{2}x^2-6x+8ln\left | x \right |+C$

$\frac{7}{2}x^2-4x+8ln\left | x \right |+C$

$\frac{5}{2}x^2-4x+8ln\left | x \right |$

$\frac{5}{2}x^2-4x+8ln\left | 2x \right |+C$

Correct answer:

$\frac{5}{2}x^2-4x+8ln\left | x \right |+C$

Explanation:

First, chop up the fraction into three separate terms:

$\int 5x-4+\frac{8}{x}dx$

Now, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{5}{2}x^2-4x+8ln\left | x \right |$

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

$\frac{5}{2}x^2-4x+8ln\left | x \right |+C$

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Example Question #781 : Integrals

$\int x^{\frac{2}{3}}dx$

Possible Answers:

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

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

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

$\frac{1}{5}x^{\frac{5}{3}}+C$

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

Correct answer:

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

Explanation:

Remember that when integrating, raise the exponent by 1 and also put that result on the denominator:

$\frac{x^{\frac{5}{3}}}{\frac{5}{3}}$

Simplify to get:

$\frac{3}{5}x^{\frac{5}{3}}$

Add C because it is an indefinite integral:

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

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Example Question #151 : Indefinite Integrals

Find $\int sin^{-1}xdx$ .

Possible Answers:

$xsin^{-1}x+C$

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

ln(sinx)+C

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

Correct answer:

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

Explanation:

Using integration by parts, let u=sin^-1u , $du=\frac{dx}{\sqrt{1-x^2}}$ , dv=dx , and v=x .

Use the formula to get $f(x)=xsin^{-1}x-\frac{1}{2}\int\frac{2x}{\sqrt{1-x^2}}$ .

Then, you can use substitution for the integral in the new equation. If f(t)=1-x^2 , then dt=-2xdx .

Using substitution, the second half of the expression becomes, $-\int \frac{dt}{\sqrt{t}}$ .

If you use the Power Rule, $f(x)=-\int t^{-\frac{1}{2}}dt=-2t^\frac{1}{2}+C.$

If you substitute this back in, you get $xsin^{-1}x-\frac{1}{2}(-2)\sqrt{1-x^2}+C$ , and you can simplify this to $xsin^{-1}x+\sqrt{1-x^2}+C$ .

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Example Question #151 : Indefinite Integrals

Solve $\int \sqrt[3]{x} + 10\sqrt[5]{x^3} dx$ .

Possible Answers:

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

$\frac{3}{4}\sqrt[3]{x^4}+(\frac{25}{4})\sqrt[5]{x^8}+C$

$\frac{3}{2}x^\frac{2}{3}+\frac{6}{3}x^\frac{1}{2}+C$

$3\sqrt[3]{x}+6\sqrt[3]{x}+C$

Correct answer:

$\frac{3}{4}\sqrt[3]{x^4}+(\frac{25}{4})\sqrt[5]{x^8}+C$

Explanation:

Splitting up the integral with the Sum Rule turns the problem into: $\int \sqrt[3]{x} + 10\sqrt[5]{x^3} dx=\int \sqrt[3]{x}+\int 10\sqrt[5]{x^3}=\int x^\frac{1}{3} + 10\int x^\frac{3}{5}$ .

We can then use the Power Rule to solve the equation:

$\frac{3}{4}x^\frac{4}{3}+10(\frac{5}{8})x^\frac{8}{5}+C$ , which can be simplified to $\frac{3}{4}\sqrt[3]{x^4}+(\frac{25}{4})\sqrt[5]{x^8}+C$ .

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Example Question #431 : Finding Integrals

Find: $\int (x^2-1)(4+3x)dx$

Possible Answers:

2x+3

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

$\frac{1}{3}x^3+\frac{3}{2}x^2+3x+C$

3x^2+2x-3

Correct answer:

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

Explanation:

First, FOIL the products so we have: $\int 4x^2+3x^3-4-3x=\int 3x^3+4x^2-3x-4$ .

Then, we can just use the Power Rule:

$\frac{3}{4}x^4+\frac{4}{3}x^3-\frac{3}{2}x^2-4x$ .

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Example Question #432 : Finding Integrals

Solve: $\int 2 cos (x)-sec(x)tan(x)dx$ .

Possible Answers:

$2\sqrt{sin(x)}-tan(x)+C$

2cos(x)-sec(x)+C

Can't be solved

2sin (x) -sec (x)+C

$2\sqrt{cos(x)}-cos(x)+C$

Correct answer:

2sin (x) -sec (x)+C

Explanation:

To solve the equation, one should know that the integral of cos(x) is sin(x) , and the integral of sec(x)tan(x) . Using these rules and pulling out the constant, we get $\int 2 cos (x)-sec(x)tan(x)dx=2sin (x) -sec (x)+C$ .

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Example Question #152 : Indefinite Integrals

Solve the indefinite integral.

$\int 2e^{2x}\,dx$

Possible Answers:

$4e^{2x}+c$

$2e^{2x}+c$

$\frac{1}{2}e^{2x}+c$

$e^{2x}+c$

Correct answer:

$e^{2x}+c$

Explanation:

To solve the indefinite integral

$\int 2e^{2x}\,dx$

we use u-substitution, setting u=2x . We derive that equation to get du=2dx and so the integral becomes

$\int e^u\,du$

The integral then becomes

e^u+c

and substituting back in for yields

$e^{2x}+c$

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Example Question #432 : Finding Integrals

Solve the indefinite integral.

$\int sec^2x\,dx$

Possible Answers:

$sin\,x+c$

$cos\,x+c$

$tan\,x+c$

$cot\,x+c$

Correct answer:

$tan\,x+c$

Explanation:

The antiderivative of sec^2x is $tan\,x$ .

As such,
$\int sec^2x\,dx=tan\,x+c$

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Example Question #432 : Finding Integrals

$\int x^3+2x$

Possible Answers:

$\frac {x^4}{4}+x^2+C$

$\frac {x^4}{4}-x^2$

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

None of the Above

Correct answer:

$\frac {x^4}{4}+x^2+C$

Explanation:

Step 1: We will first take the antiderivative of x^3 .. To do this, we add to the exponent. We then divide the entire term by the new exponent..

So, $x^3=\frac {x^{3+1}}{3+1}=\frac {x^4}{4}$

Step 2: We will take the antiderivative of 2x. We will apply the same rules that we used in Step 1...

$2x=2 \cdot \frac {x^{1+1}}{1+1}=2 \cdot \frac {x^2}{2}=x^2$

We will add the antiderivatives from both steps together..

(Recall that when taking the indefinite integral of a function, +C must be added to the integrated function as it represents any constants that may be present.)

The final answer is:

$\frac {x^4}{4}+x^2+C$

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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #431 : Finding Integrals

Example Question #432 : Finding Integrals

Example Question #781 : Integrals

Example Question #151 : Indefinite Integrals

Example Question #151 : Indefinite Integrals

Example Question #431 : Finding Integrals

Example Question #432 : Finding Integrals

Example Question #152 : Indefinite Integrals

Example Question #432 : Finding Integrals

Example Question #432 : Finding Integrals

All Calculus 2 Resources

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