∫((x+1)/(x^2 −((1+(√5))/2)x+1))dx


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Question Number 93587 by M±th+et+s last updated on 13/May/20

$\int \frac{x + 1}{x^{2} - \frac{1 + \sqrt{5}}{2} x + 1} d x$
	Answered by niroj last updated on 13/May/20
	$I= ∫ (( x+1)/(x^2 −((1+(√5))/2)x+1))dx = ∫ (((1/2)(2x−((1+(√5))/2))+ ((5+(√5))/4))/(x^2 −((1+(√5))/2)x+1))dx = (1/2)∫ (( 2x−((1+(√5))/2))/(x^2 −((1+(√5))/2)x+1))dx+ ((5+(√5))/4)∫ (( dx)/(x^2 − ((1+(√5))/2)+1)) = (1/2)log (x^2 −((1+(√5))/2)+1) + ((5+(√5))/4) ∫ (( 1)/((x)^2 −2.x.((1+(√5))/4)+( ((1+(√5))/4))^2 −(((1+(√5))/4))^2 +1))dx +C = (1/2)log (x^2 −((1+(√5))/2)+1)+ ((5+(√5))/4)∫(( 1)/((x−((1+(√5))/4))^2 −{(((1+(√5) )^2 −16)/(16))})) = (1/2)log (x^2 −((1+(√5))/2)+1)+ ((5+(√5))/4)∫ (( 1)/((x−((1+(√5))/4))^2 −((√(( (√5) −5)/8)) )^2 ))dx = (1/2)log (x^2 −((1+(√5))/2)+1)+((5+(√5))/4).(1/(2.(√(( (√5)−5)/8))))log ((x−((1+(√5))/4) −(√(( (√5)−5)/8)))/(x−((1+(√5))/4)+ (√(( (√5) −5)/8)))) +C = (1/2)log (x^2 − ((1+(√5))/2)+1)+((5+(√5))/4).(1/(2((√((√5) −5))/(2(√2)))))log ((((4x−1−(√5))/4) − ((√((√5) −5))/(2(√2))))/(((4x−1−(√5))/4) +((√((√5) −5))/(2(√2))))) +C = (1/2)log (x^2 −((1+(√5))/2)+1)+(((5+(√5) )(√2))/(4(√((√5) −5))))log (((4x−1−(√5) )(√2) −2(√((√5) −5)))/((4x−1−(√5) )(√2) +2(√((√5) −5)))) +C //.$
	$I = \int \frac{x + 1}{x^{2} - \frac{1 + \sqrt{5}}{2} x + 1} dx$ $= \int \frac{\frac{1}{2} (2 x - \frac{1 + \sqrt{5}}{2}) + \frac{5 + \sqrt{5}}{4}}{x^{2} - \frac{1 + \sqrt{5}}{2} x + 1} dx$ $= \frac{1}{2} \int \frac{2 x - \frac{1 + \sqrt{5}}{2}}{x^{2} - \frac{1 + \sqrt{5}}{2} x + 1} dx + \frac{5 + \sqrt{5}}{4} \int \frac{dx}{x^{2} - \frac{1 + \sqrt{5}}{2} + 1}$ $= \frac{1}{2} \log (x^{2} - \frac{1 + \sqrt{5}}{2} + 1) + \frac{5 + \sqrt{5}}{4} \int \frac{1}{{(x)}^{2} - 2 . x . \frac{1 + \sqrt{5}}{4} + {(\frac{1 + \sqrt{5}}{4})}^{2} - {(\frac{1 + \sqrt{5}}{4})}^{2} + 1} dx + C$ $= \frac{1}{2} \log (x^{2} - \frac{1 + \sqrt{5}}{2} + 1) + \frac{5 + \sqrt{5}}{4} \int \frac{1}{{(x - \frac{1 + \sqrt{5}}{4})}^{2} - {\frac{{(1 + \sqrt{5})}^{2} - 16}{16}}}$ $= \frac{1}{2} \log (x^{2} - \frac{1 + \sqrt{5}}{2} + 1) + \frac{5 + \sqrt{5}}{4} \int \frac{1}{{(x - \frac{1 + \sqrt{5}}{4})}^{2} - {(\sqrt{\frac{\sqrt{5} - 5}{8}})}^{2}} dx$ $= \frac{1}{2} \log (x^{2} - \frac{1 + \sqrt{5}}{2} + 1) + \frac{5 + \sqrt{5}}{4} . \frac{1}{2 . \sqrt{\frac{\sqrt{5} - 5}{8}}} \log \frac{x - \frac{1 + \sqrt{5}}{4} - \sqrt{\frac{\sqrt{5} - 5}{8}}}{x - \frac{1 + \sqrt{5}}{4} + \sqrt{\frac{\sqrt{5} - 5}{8}}} + C$ $= \frac{1}{2} \log (x^{2} - \frac{1 + \sqrt{5}}{2} + 1) + \frac{5 + \sqrt{5}}{4} . \frac{1}{2 \frac{\sqrt{\sqrt{5} - 5}}{2 \sqrt{2}}} \log \frac{\frac{4 x - 1 - \sqrt{5}}{4} - \frac{\sqrt{\sqrt{5} - 5}}{2 \sqrt{2}}}{\frac{4 x - 1 - \sqrt{5}}{4} + \frac{\sqrt{\sqrt{5} - 5}}{2 \sqrt{2}}} + C$ $= \frac{1}{2} \log (x^{2} - \frac{1 + \sqrt{5}}{2} + 1) + \frac{(5 + \sqrt{5}) \sqrt{2}}{4 \sqrt{\sqrt{5} - 5}} \log \frac{(4 x - 1 - \sqrt{5}) \sqrt{2} - 2 \sqrt{\sqrt{5} - 5}}{(4 x - 1 - \sqrt{5}) \sqrt{2} + 2 \sqrt{\sqrt{5} - 5}} + C / / .$
	Commented by M±th+et+s last updated on 13/May/20

	$g r e a t w o r k t h a n x <$
	Commented by niroj last updated on 14/May/20
	��
	Answered by Ar Brandon last updated on 14/May/20

	$K = \int \frac{x + 1}{x^{2} - \frac{1 + \sqrt{5}}{2} x + 1} d x = \frac{1}{2} \int \frac{2 x - \frac{1 + \sqrt{5}}{2}}{x^{2} - \frac{1 + \sqrt{5}}{2} x + 1} dx + \int \frac{1 + \frac{1 + \sqrt{5}}{4}}{x^{2} - \frac{1 + \sqrt{5}}{2} x + 1} dx$ $\Rightarrow K = \frac{1}{2} \ln (x^{2} - 2 x \cos 36 ° + 1) + \int \frac{1 + {\cos 36}^{°}}{{(x - {\cos 36}^{°})}^{2} + \sin^{2} 36^{°}} dx$ $\Rightarrow K = \frac{1}{2} \ln (x^{2} - 2 x \cos 36 ° + 1) + \frac{1 + {\cos 36}^{°}}{\sin^{2} 36^{°}} \int \frac{1}{{[\frac{x - {\cos 36}^{°}}{{\sin 36}^{°}}]}^{2} + 1} d x$ $\Rightarrow K = \frac{1}{2} \ln (x^{2} - 2 x \cos 36 ° + 1) + \frac{1 + {\cos 36}^{°}}{{\sin 36}^{°}} \arctan [\frac{x - {\cos 36}^{°}}{{\sin 36}^{°}}] + C$
	Commented by M±th+et+s last updated on 14/May/20

	$t h a n k y o u s i r$
	Commented by Ar Brandon last updated on 14/May/20
	You're welcome ��
	Answered by 1549442205 last updated on 14/May/20

	$F (x) = \frac{1}{2} \int \frac{(2 x - \frac{1 + \sqrt{5}}{2}) + 2 + \frac{1 + \sqrt{5}}{2}}{x^{2} - \frac{1 + \sqrt{5}}{2} x + 1} dx$ $= 0.5 \int \frac{du}{u} + (1 + \frac{1 + \sqrt{5}}{4}) \int \frac{dx}{x^{2} - \frac{1 + \sqrt{5}}{2} x + 1}$ $= 0 . 5 \ln ∣ x^{2} - \frac{1 + \sqrt{5}}{2} + 1 ∣ + \frac{5 + \sqrt{5}}{4} \int \frac{d (x - \frac{1 + \sqrt{5}}{4})}{{(x - \frac{1 + \sqrt{5}}{4})}^{2} + \frac{10 - 2 \sqrt{5}}{16}}$ $= 0 . 5 \ln ∣ x^{2} - \frac{1 + \sqrt{5}}{2} x + 1 ∣ + \frac{5 + \sqrt{5}}{4} \times \frac{4}{\sqrt{10 - 2 \sqrt{5}}} \tan^{- 1} (\frac{x - \frac{1 + \sqrt{5}}{4}}{\frac{\sqrt{10 - 2 \sqrt{5}}}{4}})$ $= 0 . 5 \ln ∣ x^{2} - \frac{1 + \sqrt{5}}{2} x + 1 ∣ + \frac{5 + \sqrt{5}}{\sqrt{10 - 2 \sqrt{5}}} \tan^{- 1} (\frac{4 x - 1 - \sqrt{5}}{\sqrt{10 - 2 \sqrt{5}}}) + C$ $=$
	Commented by M±th+et+s last updated on 14/May/20

	$t h a n k y o u s i r$
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