∫(((x−1)/(x^2 −x−1)))dx


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Question Number 92025 by 675480065 last updated on 04/May/20

$\int (\frac{x - 1}{x^{2} - x - 1}) dx$
Commented by mathmax by abdo last updated on 04/May/20
$I =∫ ((x−1)/(x^2 −x−1))dx ⇒ I =(1/2)∫((2x−1−1)/(x^2 −x−1))dx =(1/2)∫ ((2x−1)/(x^2 −x−1))dx −(1/2)∫ (dx/(x^2 −x−1)) we have ∫ ((2x−1)/(x^2 −x−1))dx =ln∣x^2 −x−1∣ +c_0 x^2 −x−1=0→Δ =1+4 =5 ⇒x_1 =((1+(√5))/2) and x_2 =((1−(√5))/2) ⇒ ∫ (dx/(x^2 −x−1)) =∫(dx/((x−((1+(√5))/2))(x−((1−(√5))/2)))) =∫ (1/(√5)) {(1/(x−((1+(√5))/2)))−(1/(x−((1−(√5))/2)))}dx =(1/(√5))ln∣((x−((1+(√5))/2))/(x−((1−(√5))/2)))∣ +c_1 =(1/(√5))ln∣((2x−1−(√5))/(2x−1+(√5)))∣ +c_1 ⇒ I =(1/2)ln∣x^2 −x−1∣−(1/(2(√5)))ln∣((2x−1−(√5))/(2x−1+(√5)))∣ +C$
$I = \int \frac{x - 1}{x^{2} - x - 1} d x \Rightarrow I = \frac{1}{2} \int \frac{2 x - 1 - 1}{x^{2} - x - 1} d x$ $= \frac{1}{2} \int \frac{2 x - 1}{x^{2} - x - 1} d x - \frac{1}{2} \int \frac{d x}{x^{2} - x - 1} w e h a v e$ $\int \frac{2 x - 1}{x^{2} - x - 1} d x = l n ∣ x^{2} - x - 1 ∣ + c_{0}$ $x^{2} - x - 1 = 0 \to Δ = 1 + 4 = 5 \Rightarrow x_{1} = \frac{1 + \sqrt{5}}{2} a n d x_{2} = \frac{1 - \sqrt{5}}{2} \Rightarrow$ $\int \frac{d x}{x^{2} - x - 1} = \int \frac{d x}{(x - \frac{1 + \sqrt{5}}{2}) (x - \frac{1 - \sqrt{5}}{2})}$ $= \int \frac{1}{\sqrt{5}} {\frac{1}{x - \frac{1 + \sqrt{5}}{2}} - \frac{1}{x - \frac{1 - \sqrt{5}}{2}}} d x$ $= \frac{1}{\sqrt{5}} l n ∣ \frac{x - \frac{1 + \sqrt{5}}{2}}{x - \frac{1 - \sqrt{5}}{2}} ∣ + c_{1} = \frac{1}{\sqrt{5}} l n ∣ \frac{2 x - 1 - \sqrt{5}}{2 x - 1 + \sqrt{5}} ∣ + c_{1} \Rightarrow$ $I = \frac{1}{2} l n ∣ x^{2} - x - 1 ∣ - \frac{1}{2 \sqrt{5}} l n ∣ \frac{2 x - 1 - \sqrt{5}}{2 x - 1 + \sqrt{5}} ∣ + C$
	Answered by niroj last updated on 04/May/20

	$\int \frac{x - 1}{x^{2} - x - 1} dx = \frac{1}{2} \int \frac{2 x - 1 - 1}{x^{2} - x - 1} dx$ $= \frac{1}{2} \int \frac{2 x - 1}{x^{2} - x - 1} dx - \frac{1}{2} \int \frac{1}{x^{2} - x - 1} dx$ $= \frac{1}{2} \log (x^{2} - x - 1) + \frac{1}{2} \int \frac{1}{x^{2} - 2 x . \frac{1}{2} + \frac{1}{4} - \frac{1}{4} - 1} dx + C$ $= \log {(x^{2} - x - 1)}^{\frac{1}{2}} + \frac{1}{2} \int \frac{1}{{(x - \frac{1}{2})}^{2} - {(\frac{\sqrt{5}}{2})}^{2}} dx + C$ $= \log \sqrt{x^{2} - x - 1} + \frac{1}{2} [\frac{1}{2 . \frac{\sqrt{5}}{2}} \log \frac{x - \frac{1}{2} - \frac{\sqrt{5}}{2}}{x - \frac{1}{2} + \frac{\sqrt{5}}{2}}] + C$ $= \log \sqrt{x^{2} - x - 1} + \frac{1}{2 \sqrt{5}} \log \frac{2 x - 1 - \sqrt{5}}{2 x - 1 + \sqrt{5}} + C$ $= \log \sqrt{x^{2} - x - 1} + \frac{\sqrt{5}}{10} \log \frac{2 x - 1 - \sqrt{5}}{2 x - 1 + \sqrt{5}} + C / / .$
	Answered by MJS last updated on 04/May/20

	$\int \frac{x - 1}{x^{2} - x - 1} d x =$ $= \int \frac{5 + \sqrt{5}}{5 (2 x - 1 + \sqrt{5})} d x + \int \frac{5 - \sqrt{5}}{5 (2 x - 1 - \sqrt{5})} d x =$ $= \frac{5 + \sqrt{5}}{10} \ln (2 x - 1 + \sqrt{5}) + \frac{5 - \sqrt{5}}{10} \ln (2 x - 1 - \sqrt{5}) + C$
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