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Question Number 51510 by Tinkutara last updated on 27/Dec/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 28/Dec/18
	$∫((−2pdp)/(√(p^2 ×({1−3(p^2 −1)}))) x^2 {x^2 +(1/x^2 )+2(x+(1/x))−1} x^2 {(x+(1/x))^2 −2+2(x+(1/x))−1} x^2 {(x+(1/x))^2 +2(x+(1/x))−3} now ∫(((x−1)x(√({(x+(1/x))^2 +2(x+(1/x))−3))[)/(x^2 (x+1)))dx ∫(((x−1)(√({(x+(1/x))^2 +2(x+(1/x))−3))[)/(x^2 +x))dx ∫(((x^2 −1)(√({(x+(1/x))^2 +2(x+(1/x))−3)) )/(x(x+1)^2 ))dx ∫(((x^2 −1)(√({(x+(1/x))^2 +2(x+(1/x))−3)))/(x(x^2 +2x+1)))dx ∫(((1−(1/x^2 ))(√({(x+(1/x))^2 +2(x+(1/x))−3)))/((x+(1/x)+2)))dx put k=x+(1/x) dk=(1−(1/x^2 ))dx ∫((√(k^2 +2k−3))/(k+2)) dk wait i am near solution... ∫((k(k+2)−3)/((k+2)(√(k^2 +2k−3)) ))dk ∫(k/(√(k^2 +2k−3)))dk−3∫(dk/((k+2)(√(k^2 +2k−3)))) ∫((k+1−1)/(√(k^2 +2k−3)))dk−3∫(dk/((k+2)(√((k^2 +2k−3))))) (1/2)∫((d(k^2 +2k−3))/(√(k^2 +2k−3)))−∫(dk/(√((k+1)^2 −2^2 )))−3∫(dk/((k+2)(√(k^2 +2k−3)))) I=(1/2)I_1 −I_2 −3I_3 I_1 =(((k^2 +2k−3)^(((−1)/2)+1) )/(((−1)/2)+1))=2(k^2 +2k−3)^(1/2) I_2 =∫(dk/(√((k+1)^2 −2^2 )))=ln{(k+1)+(√((k+1)^2 −2^2 )) } I_2 =ln{(k+1)+(√(k^2 +2k−3)) } I_3 =∫(dk/((k+2)(√(k^2 +2k−3))))→∫(dk/((k+2)(√((k+3)(k−1))))) k+2=(1/t) dk=((−1)/t^2 )dt ∫(((−1)/t^2 )/((1/t)(√(((1/t)+1)((1/t)−3)))))dt ∫(((−1)/t)/((√((1+t)(1−3t))) ))×tdt =∫((−dt)/(√((1+t)(1−3t)))) =∫((−dt)/(√(1−2t−3t^2 ))) calculation −3t^2 −2t+1 =−3(t^2 +((2t)/3)+(1/(9 ))−(1/9)−(1/3)) =−3{(t+(1/3))^2 −((2/3))^2 }=3{((2/3))^2 −(t+(1/3))^2 } ∫((−dt)/(√(3{((2/3))^2 −(t+(1/3))^2 }))) =((−1)/(√3))sin^(−1) (((t+(1/3))/(2/3)))=((−1)/(√3))sin^(−1) (((3t+1)/2))=((−1)/(√3))sin^(−1) (((3((1/(k+2)))+1)/2)) =((−1)/(√3))sin^(−1) (((k+5)/(2k+4))) so answer is (1/2)×2(k^2 +2k−3)^(1/2) −ln{(k+1)+(√(k^2 +2k−3)) )}−3×(((−1)/(√3)))sin^(−1) (((k+5)/(2k+4))) =(k^2 +2k−3)^(1/2) −ln{(k+1)+(√(k^2 +2k−3)) )}+(√3) sin^(−1) (((k+5)/(2k+4)))$
	$\int \frac{- 2 p d p}{\sqrt{p^{2} \times ({1 - 3 (p^{2} - 1)}}}$ $x^{2} {x^{2} + \frac{1}{x^{2}} + 2 (x + \frac{1}{x}) - 1}$ $x^{2} {{(x + \frac{1}{x})}^{2} - 2 + 2 (x + \frac{1}{x}) - 1}$ $x^{2} {{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3}$ $n o w$ $\int \frac{(x - 1) x \sqrt{{{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3} [}{x^{2} (x + 1)} d x$ $\int \frac{(x - 1) \sqrt{{{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3} [}{x^{2} + x} d x$ $\int \frac{(x^{2} - 1) \sqrt{{{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3}}{x {(x + 1)}^{2}} d x$ $\int \frac{(x^{2} - 1) \sqrt{{{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3}}{x (x^{2} + 2 x + 1)} d x$ $\int \frac{(1 - \frac{1}{x^{2}}) \sqrt{{{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3}}{(x + \frac{1}{x} + 2)} d x$ $p u t k = x + \frac{1}{x} d k = (1 - \frac{1}{x^{2}}) d x$ $\int \frac{\sqrt{k^{2} + 2 k - 3}}{k + 2} d k$ $w a i t i a m n e a r s o l u t i o n . . .$ $\int \frac{k (k + 2) - 3}{(k + 2) \sqrt{k^{2} + 2 k - 3}} d k$ $\int \frac{k}{\sqrt{k^{2} + 2 k - 3}} d k - 3 \int \frac{d k}{(k + 2) \sqrt{k^{2} + 2 k - 3}}$ $\int \frac{k + 1 - 1}{\sqrt{k^{2} + 2 k - 3}} d k - 3 \int \frac{d k}{(k + 2) \sqrt{(k^{2} + 2 k - 3)}}$ $\frac{1}{2} \int \frac{d (k^{2} + 2 k - 3)}{\sqrt{k^{2} + 2 k - 3}} - \int \frac{d k}{\sqrt{{(k + 1)}^{2} - 2^{2}}} - 3 \int \frac{d k}{(k + 2) \sqrt{k^{2} + 2 k - 3}}$ $I = \frac{1}{2} I_{1} - I_{2} - 3 I_{3}$ $I_{1} = \frac{{(k^{2} + 2 k - 3)}^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} = 2 {(k^{2} + 2 k - 3)}^{\frac{1}{2}}$ $I_{2} = \int \frac{d k}{\sqrt{{(k + 1)}^{2} - 2^{2}}} = l n {(k + 1) + \sqrt{{(k + 1)}^{2} - 2^{2}}}$ $I_{2} = l n {(k + 1) + \sqrt{k^{2} + 2 k - 3}}$ $I_{3} = \int \frac{d k}{(k + 2) \sqrt{k^{2} + 2 k - 3}} \to \int \frac{d k}{(k + 2) \sqrt{(k + 3) (k - 1)}}$ $k + 2 = \frac{1}{t} d k = \frac{- 1}{t^{2}} d t$ $\int \frac{\frac{- 1}{t^{2}}}{\frac{1}{t} \sqrt{(\frac{1}{t} + 1) (\frac{1}{t} - 3)}} d t$ $\int \frac{\frac{- 1}{t}}{\sqrt{(1 + t) (1 - 3 t)}} \times t d t$ $= \int \frac{- d t}{\sqrt{(1 + t) (1 - 3 t)}} = \int \frac{- d t}{\sqrt{1 - 2 t - 3 t^{2}}}$ $c a l c u l a t i o n$ $- 3 t^{2} - 2 t + 1$ $= - 3 (t^{2} + \frac{2 t}{3} + \frac{1}{9} - \frac{1}{9} - \frac{1}{3})$ $= - 3 {{(t + \frac{1}{3})}^{2} - {(\frac{2}{3})}^{2}} = 3 {{(\frac{2}{3})}^{2} - {(t + \frac{1}{3})}^{2}}$ $\int \frac{- d t}{\sqrt{3 {{(\frac{2}{3})}^{2} - {(t + \frac{1}{3})}^{2}}}}$ $= \frac{- 1}{\sqrt{3}} {s i n}^{- 1} (\frac{t + \frac{1}{3}}{\frac{2}{3}}) = \frac{- 1}{\sqrt{3}} {s i n}^{- 1} (\frac{3 t + 1}{2}) = \frac{- 1}{\sqrt{3}} {s i n}^{- 1} (\frac{3 (\frac{1}{k + 2}) + 1}{2})$ $= \frac{- 1}{\sqrt{3}} {s i n}^{- 1} (\frac{k + 5}{2 k + 4})$ $s o a n s w e r i s$ $\frac{1}{2} \times 2 {(k^{2} + 2 k - 3)}^{\frac{1}{2}} - l n {(k + 1) + \sqrt{k^{2} + 2 k - 3})} - 3 \times (\frac{- 1}{\sqrt{3}}) {s i n}^{- 1} (\frac{k + 5}{2 k + 4})$ $= {(k^{2} + 2 k - 3)}^{\frac{1}{2}} - l n {(k + 1) + \sqrt{k^{2} + 2 k - 3})} + \sqrt{3} {s i n}^{- 1} (\frac{k + 5}{2 k + 4})$
	Commented by Tinkutara last updated on 27/Dec/18

	Commented by Tinkutara last updated on 28/Dec/18
	Yes Sir, I am also getting the same answer as yours. There should be +√3 sin inverse…
	Commented by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18

	$i n w h i t e p a p e r i t c a n b e s o l v e d q u i c k l y$ $n o n e e d o f t y p i n g . .$ $i h a v e s o l v e d t h e l e n g t h y p r o b l e m$ $\int \frac{\sqrt{k^{2} + 2 k - 3}}{k + 2} d k . . . p l s c h e c k . . .$
	Commented by Tinkutara last updated on 27/Dec/18
	This is the answer given.
	Commented by tanmay.chaudhury50@gmail.com last updated on 28/Dec/18

	$p l s c h e c k . . . f i n a l a n s s e r . . .$
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