Respuesta :
Answer:
[tex]\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[/tex]
General Formulas and Concepts:
Pre-Algebra
- Distributive Property
Algebra I
- Completing the Square
- Rearranging Variables
Algebra II
- Long Division
Calculus
- U-Substitution
- [Integration Trick 1] Numerator Split
- [Integration Trick 2] Completing the Square
- Integration Rule 1: [tex]\int {cf(x)} \, dx = c\int {f(x)} \, dx[/tex]
- Integration Rule 2: [tex]\int {f(x)+g(x)} \, dx =\int {f(x)} \, dx + \int {g(x)} \, dx[/tex]
- Integration 1: [tex]\int {\frac{1}{u} } \, du =ln|u| + C[/tex]
- Integration 2: [tex]\int {\frac{du}{u^2+a^2} } = \frac{1}{a} arctan(\frac{u}{a} )+C[/tex]
- Integration 3: [tex]\int {x^n} \, dx = \frac{x^{n+1}}{n+1} +C[/tex]
Step-by-step explanation:
Step 1: Define
[tex]\int {\frac{x^2}{x^2+x+3} } \, dx[/tex]
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 [tex]1-\frac{x+3}{x^2+x+3}[/tex]
Now we rewrite our Integral: [tex]\int ({1-\frac{x+3}{x^2+x+3} }) \, dx[/tex]
Step 3: Integrate Pt. 1
- Distributive Integral [Int Rule 1]: [tex]\int {1} \, dx - \int {\frac{x+3}{x^2+x+3} } \, dx[/tex]
- Integrate 1st Integral [Int 3]: [tex]x - \int {\frac{x+3}{x^2+x+3} } \, dx[/tex]
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]: [tex]x - \frac{1}{2} \int {\frac{2(x+3)}{x^2+x+3} } \, dx[/tex]
- Distribute 2 [Alg]: [tex]x - \frac{1}{2} \int {\frac{2x+6}{x^2+x+3} } \, dx[/tex]
- Rewrite Integral [Alg]: [tex]x - \frac{1}{2} \int {\frac{2x+1+5}{x^2+x+3} } \, dx[/tex]
- Rewrite Integral [Int Trick 1]: [tex]x - \frac{1}{2} [\int {\frac{2x+1}{x^2+x+3} } \, dx + \int {\frac{5}{x^2+x+3} } \, dx ][/tex]
- (2nd Int) Complete the Square: [tex]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 ][/tex]
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: [tex]x - \frac{1}{2} [\int {\frac{1}{u} } \, du + \int {\frac{5}{z^2 + (\sqrt{\frac{11}{4}})^2} } \, dz ][/tex]
- Rewrite Integral [Int Rule 1]: [tex]x - \frac{1}{2} [\int {\frac{1}{u} } \, du + 5\int {\frac{dz}{z^2 + (\sqrt{\frac{11}{4}})^2} } ][/tex]
- Integrate 1st Integral [Int 1]: [tex]x - \frac{1}{2} [ln|u| + 5\int {\frac{dz}{z^2 + (\sqrt{\frac{11}{4}})^2} } ][/tex]
- Integrate 2nd Integral [Int 2]: [tex]x - \frac{1}{2} [ln|u| + 5(\frac{1}{\sqrt{\frac{11}{4}}}arctan(\frac{z}{\sqrt{\frac{11}{4} } } ) ) ][/tex]
- Distribute 5 [Alg]: [tex]x - \frac{1}{2} [ln|u| + \frac{5}{\sqrt{\frac{11}{4}}}arctan(\frac{z}{\sqrt{\frac{11}{4} } } ) ][/tex]
- Distribute -1/2 [Alg]: [tex]x - \frac{1}{2}ln|u| - \frac{5}{2\sqrt{\frac{11}{4}}}arctan(\frac{z}{\sqrt{\frac{11}{4} } } )[/tex]
- Rationalize [Alg]: [tex]x - \frac{1}{2}ln|u| - \frac{5\sqrt{11} }{11}arctan(\frac{z}{\sqrt{\frac{11}{4} } } )[/tex]
- Resubstitute variables [Alg]: [tex]x - \frac{1}{2}ln|x^2+x+3| - \frac{5\sqrt{11} }{11}arctan(\frac{x+\frac{1}{2} }{\sqrt{\frac{11}{4} } } )[/tex]
- Simplify/Rationalize [Alg]: [tex]x - \frac{1}{2}ln|x^2+x+3| - \frac{5\sqrt{11} }{11}arctan(\frac{\sqrt{11}(2x+1) }{11} )[/tex]
- Rewrite [Alg]: [tex]- \frac{5\sqrt{11} }{11}arctan(\frac{\sqrt{11}(2x+1) }{11} ) - \frac{1}{2}ln|x^2+x+3| +x[/tex]
- Integration Constant: [tex]- \frac{5\sqrt{11} }{11}arctan(\frac{\sqrt{11}(2x+1) }{11} ) - \frac{1}{2}ln|x^2+x+3| +x + C[/tex]
And we have our final answer! Hope this helped you on your Calculus Journey!
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