Answer:
The answer is 8/12
Step-by-step explanation:
So, in this case you would have to add the nemeratour and the denominator. Hope this helps! :-)
Answer: You are smart, figure it out yourself
Step-by-step explanation:
Answer:
24%
i think
Step-by-step explanation:
Answer:
![\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=-\dfrac{2}{x+1}+\text{C}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B2%7D%7Bx%5E2%2B2x%2B1%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%3D-%5Cdfrac%7B2%7D%7Bx%2B1%7D%2B%5Ctext%7BC%7D)
Step-by-step explanation:
<u>Fundamental Theorem of Calculus</u>
![\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Ctext%7Bf%7D%28x%29%5C%3A%5Ctext%7Bd%7Dx%3D%5Ctext%7BF%7D%28x%29%2B%5Ctext%7BC%7D%20%5Ciff%20%5Ctext%7Bf%7D%28x%29%3D%5Cdfrac%7B%5Ctext%7Bd%7D%7D%7B%5Ctext%7Bd%7Dx%7D%28%5Ctext%7BF%7D%28x%29%29)
If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a <u>constant of integration</u>.
<u>Given integral</u>:
![\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B2%7D%7Bx%5E2%2B2x%2B1%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
Factor the denominator:
![\begin{aligned}\implies x^2+2x+1 & = x^2+x+x+1\\& = x(x+1)+1(x+1)\\& = (x+1)(x+1)\\& = (x+1)^2\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cimplies%20x%5E2%2B2x%2B1%20%26%20%3D%20x%5E2%2Bx%2Bx%2B1%5C%5C%26%20%3D%20x%28x%2B1%29%2B1%28x%2B1%29%5C%5C%26%20%3D%20%20%28x%2B1%29%28x%2B1%29%5C%5C%26%20%3D%20%20%28x%2B1%29%5E2%5Cend%7Baligned%7D)
![\implies \displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=\int \dfrac{2}{(x+1)^2}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B2%7D%7Bx%5E2%2B2x%2B1%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%3D%5Cint%20%5Cdfrac%7B2%7D%7B%28x%2B1%29%5E2%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
![\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%20%5Cdfrac%7B1%7D%7Ba%5En%7D%3Da%5E%7B-n%7D)
![\implies \displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=\int 2(x+1)^{-2}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B2%7D%7Bx%5E2%2B2x%2B1%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%3D%5Cint%202%28x%2B1%29%5E%7B-2%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
![\boxed{\begin{minipage}{4 cm}\underline{Integrating $ax^n$}\\\\$\displaystyle \int ax^n\:\text{d}x=\dfrac{ax^{n+1}}{n+1}+\text{C}$\end{minipage}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cbegin%7Bminipage%7D%7B4%20cm%7D%5Cunderline%7BIntegrating%20%24ax%5En%24%7D%5C%5C%5C%5C%24%5Cdisplaystyle%20%5Cint%20ax%5En%5C%3A%5Ctext%7Bd%7Dx%3D%5Cdfrac%7Bax%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D%2B%5Ctext%7BC%7D%24%5Cend%7Bminipage%7D%7D)
Use <u>Integration by Substitution</u>:
![\textsf{Let }u=(x+1) \implies \dfrac{\text{d}u}{\text{d}x}=1 \implies \text{d}x=\text{d}u}](https://tex.z-dn.net/?f=%5Ctextsf%7BLet%20%7Du%3D%28x%2B1%29%20%5Cimplies%20%5Cdfrac%7B%5Ctext%7Bd%7Du%7D%7B%5Ctext%7Bd%7Dx%7D%3D1%20%5Cimplies%20%5Ctext%7Bd%7Dx%3D%5Ctext%7Bd%7Du%7D)
Therefore:
![\begin{aligned}\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x & = \int 2(x+1)^{-2}\:\:\text{d}x\\\\& = \int 2u^{-2}\:\:\text{d}u\\\\& = \dfrac{2}{-1}u^{-2+1}+\text{C}\\\\& = -2u^{-1}+\text{C}\\\\& = -\dfrac{2}{u}+\text{C}\\\\& = -\dfrac{2}{x+1}+\text{C}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B2%7D%7Bx%5E2%2B2x%2B1%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%3D%20%5Cint%202%28x%2B1%29%5E%7B-2%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%5C%5C%5C%5C%26%20%3D%20%5Cint%202u%5E%7B-2%7D%5C%3A%5C%3A%5Ctext%7Bd%7Du%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B2%7D%7B-1%7Du%5E%7B-2%2B1%7D%2B%5Ctext%7BC%7D%5C%5C%5C%5C%26%20%3D%20-2u%5E%7B-1%7D%2B%5Ctext%7BC%7D%5C%5C%5C%5C%26%20%3D%20-%5Cdfrac%7B2%7D%7Bu%7D%2B%5Ctext%7BC%7D%5C%5C%5C%5C%26%20%3D%20-%5Cdfrac%7B2%7D%7Bx%2B1%7D%2B%5Ctext%7BC%7D%5Cend%7Baligned%7D)
Learn more about integration here:
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