Mathematics, 16.12.2020 05:30, TheHomieJaay3092

Integrate the following:

$\displaystyle \int \frac{x^2}{x^2+x+3}\, dx$

Answers: 3

Show answers

Answers

answered: sanchezr7187

$\int {\frac{x^2}{x^2+x+3} } \, dx = - \frac{5\sqrt{11} }{11}arctan(\frac{\sqrt{11}(2x+1) }{11} ) - \frac{1}{2}ln|x^2+x+3| +x + C$

General Formulas and Concepts:

Pre-Algebra

Distributive Property

Algebra I

Completing the SquareRearranging Variables

Algebra II

Long Division

Calculus

U-Substitution[Integration Trick 1] Numerator Split[Integration Trick 2] Completing the SquareIntegration Rule 1: $\int {cf(x)} \, dx = c\int {f(x)} \, dx$ Integration Rule 2: $\int {f(x)+g(x)} \, dx =\int {f(x)} \, dx + \int {g(x)} \, dx$ Integration 1: $\int {\frac{1}{u} } \, du =ln|u| + C$ Integration 2: $\int {\frac{du}{u^2+a^2} } = \frac{1}{a} arctan(\frac{u}{a} )+C$ Integration 3: $\int {x^n} \, dx = \frac{x^{n+1}}{n+1} +C$

Step-by-step explanation:

Step 1: Define

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

Step 2: Simplify Function

We do long division to simplify the function inside the function.

See Attachment for Long Division Work.

Once we do long division, our function becomes $1-\frac{x+3}{x^2+x+3}$

Now we rewrite our Integral: $\int ({1-\frac{x+3}{x^2+x+3} }) \, dx$

Step 3: Integrate Pt. 1

Distributive Integral [Int Rule 1]: $\int {1} \, dx - \int {\frac{x+3}{x^2+x+3} } \, dx$ Integrate 1st Integral [Int 3]: $x - \int {\frac{x+3}{x^2+x+3} } \, dx$

Step 4: Identify Variables Pt.1

Set variables for u-substitution.

u = x² + x + 3

du = (2x + 1)dx

Step 5: Integrate Pt. 2

Rewrite Integral [Int Rule 1]: $x - \frac{1}{2} \int {\frac{2(x+3)}{x^2+x+3} } \, dx$ Distribute 2 [Alg]: $x - \frac{1}{2} \int {\frac{2x+6}{x^2+x+3} } \, dx$ Rewrite Integral [Alg]: $x - \frac{1}{2} \int {\frac{2x+1+5}{x^2+x+3} } \, dx$ Rewrite Integral [Int Trick 1]: $x - \frac{1}{2} [\int {\frac{2x+1}{x^2+x+3} } \, dx + \int {\frac{5}{x^2+x+3} } \, dx ]$ (2nd Int) Complete the Square: $x - \frac{1}{2} [\int {\frac{2x+1}{x^2+x+3} } \, dx + \int {\frac{5}{(x+\frac{1}{2})^2 + \frac{11}{4} } } \, dx ]$

Step 6: Identify Variables Pt. 2

Set variables for u-substitution for 2nd integral.

z = x + 1/2

dz = dx

a = √(11/4)

Step 7: Integrate Pt. 3

[Integrate] U-Substitution: $x - \frac{1}{2} [\int {\frac{1}{u} } \, du + \int {\frac{5}{z^2 + (\sqrt{\frac{11}{4}})^2} } \, dz ]$ Rewrite Integral [Int Rule 1]: $x - \frac{1}{2} [\int {\frac{1}{u} } \, du + 5\int {\frac{dz}{z^2 + (\sqrt{\frac{11}{4}})^2} } ]$ Integrate 1st Integral [Int 1]: $x - \frac{1}{2} [ln|u| + 5\int {\frac{dz}{z^2 + (\sqrt{\frac{11}{4}})^2} } ]$ Integrate 2nd Integral [Int 2]: $x - \frac{1}{2} [ln|u| + 5(\frac{1}{\sqrt{\frac{11}{4}}}arctan(\frac{z}{\sqrt{\frac{11}{4} } } ) ) ]$ Distribute 5 [Alg]: $x - \frac{1}{2} [ln|u| + \frac{5}{\sqrt{\frac{11}{4}}}arctan(\frac{z}{\sqrt{\frac{11}{4} } } ) ]$ Distribute -1/2 [Alg]: $x - \frac{1}{2}ln|u| - \frac{5}{2\sqrt{\frac{11}{4}}}arctan(\frac{z}{\sqrt{\frac{11}{4} } } )$ Rationalize [Alg]: $x - \frac{1}{2}ln|u| - \frac{5\sqrt{11} }{11}arctan(\frac{z}{\sqrt{\frac{11}{4} } } )$ Resubstitute variables [Alg]: $x - \frac{1}{2}ln|x^2+x+3| - \frac{5\sqrt{11} }{11}arctan(\frac{x+\frac{1}{2} }{\sqrt{\frac{11}{4} } } )$ Simplify/Rationalize [Alg]: $x - \frac{1}{2}ln|x^2+x+3| - \frac{5\sqrt{11} }{11}arctan(\frac{\sqrt{11}(2x+1) }{11} )$ Rewrite [Alg]: $- \frac{5\sqrt{11} }{11}arctan(\frac{\sqrt{11}(2x+1) }{11} ) - \frac{1}{2}ln|x^2+x+3| +x$ Integration Constant: $- \frac{5\sqrt{11} }{11}arctan(\frac{\sqrt{11}(2x+1) }{11} ) - \frac{1}{2}ln|x^2+x+3| +x + C$

And we have our final answer! Hope this helped you on your Calculus Journey!

answered: kayleewoodard

hope this will help you........

answered: Guest

step-by-step explanation:

this is not a question

answered: Guest

exa:

)

3+11

step-by-step explanation:

from this example, we can see that subtracting with integers is the same as adding the opposite. we can solve using a number line or by using the saying "keep, change, change" to us solve. it is important to remember that we can subtract a negative and end up with a larger number than we started with.