How do you integrate $\int \frac{2 x - 2}{{(8 + 2 x - x^{2})}^{\frac{1}{2}}} d x$ $int (2x-2)/ ((8 + 2x - x^2)^(1/2))dx$ using partial fractions?

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How do you integrate $\int \frac{2 x - 2}{{(8 + 2 x - x^{2})}^{\frac{1}{2}}} d x$ $int (2x-2)/ ((8 + 2x - x^2)^(1/2))dx$ using partial fractions?

1 Answer

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Answer 1 · 2016-11-02T20:23:03

The answer is $= - 2 \sqrt{8 + 2 x - x^{2}} + C$ $=-2sqrt(8+2x-x^2)+C$

Explanation:

You can integrate only by substitution.
Let $u = 8 + 2 x - x^{2}$ $u=8+2x-x^2$ $\Rightarrow$ $=>$ $d u = (2 - 2 x) d x = - (2 x - 2) d x$ $du=(2-2x)dx=-(2x-2)dx$
$\int \frac{(2 x - 2) d x}{\sqrt{8 + 2 x - x^{2}}} = - \int \frac{d u}{\sqrt{u}} = - \frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}$ $int((2x-2)dx)/(sqrt(8+2x-x^2))=-int(du)/sqrtu=-u^(-1/2+1)/(-1/2+1)$
$= - 2 \sqrt{u} = - 2 \sqrt{8 + 2 x - x^{2}} + C$ $=-2sqrtu=-2sqrt(8+2x-x^2)+C$

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How do you integrate $\int \frac{2 x - 2}{{(8 + 2 x - x^{2})}^{\frac{1}{2}}} d x$ $int (2x-2)/ ((8 + 2x - x^2)^(1/2))dx$ using partial fractions?

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How do you integrate $\int \frac{2 x - 2}{{(8 + 2 x - x^{2})}^{\frac{1}{2}}} d x$ $int (2x-2)/ ((8 + 2x - x^2)^(1/2))dx$ using partial fractions?