∫ (((x−1)(√(x^4 +2x^3 −x^2 +2x+1)))/(x^2 (x+1))) dx ?


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Question Number 123920 by john_santu last updated on 29/Nov/20

$\int \frac{(x - 1) \sqrt{x^{4} + 2 x^{3} - x^{2} + 2 x + 1}}{x^{2} (x + 1)} d x ?$
	Answered by liberty last updated on 29/Nov/20
	$(•) ((x−1)/(x^2 (x+1))) = ((x^2 −1)/(x^2 (x+1)^2 )) =(1−(1/x^2 ))((1/(x^2 (x+(1/x)+2)))) (••) x^4 +2x^3 −x^2 +2x+1=x^2 (x^2 +(1/x^2 )+2(x+(1/x))−1) = x^2 ((x+(1/x))^2 +2(x+(1/x))−3)=x^2 ((x+(1/x)+1)^2 −4) letting 2sec v = x+(1/x)+1 T=∫ ((√(4sec^2 v−4))/(2sec v+1)) (2sec v tan v )dv T=∫ ((4tan^2 v sec v )/(2sec v +1)) dv = ∫(2sec^2 v−sec v)dv−∫((3sec v dv)/(2sec v +1)) T= 2tan v −ℓn ∣sec v + tan v∣ −∫ ((3(2−cos v))/(4−cos^2 v)) dv let L=∫ ((6−3cos v)/(4−cos^2 v)) dv = ∫ ((6sec^2 v)/(4sec^2 v−1))dv−∫((3cos v)/(3+sin^2 v)) dv L=∫ ((6du)/(4u^2 +3)) [ u = tan v ] −∫((3dj)/(3+j^2 )) [ j = sin v ] L= (√3) tan^(−1) (((2u)/( (√3))))−(√3) tan^(−1) ((j/( (√3)))) L= (√3) tan^(−1) (((2tan v)/( (√3))))−(√3) tan^(−1) (((sin v)/( (√3)))) Thus T = 2tan v −ℓn∣sec v + tan v ∣− (√3) {tan^(−1) (((2tan v)/( (√3))))−tan^(−1) (((sin v)/( (√3))))}+ c where v = sec^(−1) (((x^2 +x+1)/(2x))) .$
	$(∙) \frac{x - 1}{x^{2} (x + 1)} = \frac{x^{2} - 1}{x^{2} {(x + 1)}^{2}} = (1 - \frac{1}{x^{2}}) (\frac{1}{x^{2} (x + \frac{1}{x} + 2)})$ $(∙ ∙) x^{4} + 2 x^{3} - x^{2} + 2 x + 1 = x^{2} (x^{2} + \frac{1}{x^{2}} + 2 (x + \frac{1}{x}) - 1)$ $= x^{2} ({(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3) = x^{2} ({(x + \frac{1}{x} + 1)}^{2} - 4)$ $l e t t i n g 2 \sec v = x + \frac{1}{x} + 1$ $T = \int \frac{\sqrt{4 \sec^{2} v - 4}}{2 \sec v + 1} (2 \sec v \tan v) d v$ $T = \int \frac{4 \tan^{2} v \sec v}{2 \sec v + 1} d v = \int (2 \sec^{2} v - \sec v) d v - \int \frac{3 \sec v d v}{2 \sec v + 1}$ $T = 2 \tan v - ℓ n ∣ \sec v + \tan v ∣ - \int \frac{3 (2 - \cos v)}{4 - \cos^{2} v} d v$ $l e t L = \int \frac{6 - 3 \cos v}{4 - \cos^{2} v} d v = \int \frac{6 \sec^{2} v}{4 \sec^{2} v - 1} d v - \int \frac{3 \cos v}{3 + \sin^{2} v} d v$ $L = \int \frac{6 d u}{4 u^{2} + 3} [u = \tan v] - \int \frac{3 d j}{3 + j^{2}} [j = \sin v]$ $L = \sqrt{3} \tan^{- 1} (\frac{2 u}{\sqrt{3}}) - \sqrt{3} \tan^{- 1} (\frac{j}{\sqrt{3}})$ $L = \sqrt{3} \tan^{- 1} (\frac{2 \tan v}{\sqrt{3}}) - \sqrt{3} \tan^{- 1} (\frac{\sin v}{\sqrt{3}})$ $T h u s T = 2 \tan v - ℓ n ∣ \sec v + \tan v ∣ -$ $\sqrt{3} {\tan^{- 1} (\frac{2 \tan v}{\sqrt{3}}) - \tan^{- 1} (\frac{\sin v}{\sqrt{3}})} + c$ $w h e r e v = \sec^{- 1} (\frac{x^{2} + x + 1}{2 x}) .$
	Answered by MJS_new last updated on 29/Nov/20

	$\int \frac{(x - 1) \sqrt{x^{4} + 2 x^{3} - x^{2} + 2 x + 1}}{x^{2} (x + 1)} d x =$ $= \int \frac{(x - 1) \sqrt{(x^{2} - x + 1) (x^{2} + 3 x + 1)}}{x^{2} (x + 1)} d x =$ $[t = \frac{\sqrt{x^{2} - x + 1}}{\sqrt{x^{2} + 3 x + 1}} \to d x = \frac{{(x^{2} + 3 x + 1)}^{3 / 2} \sqrt{x^{2} - x + 1}}{2 (x^{2} - 1)} d t]$ $= 32 \int \frac{t^{2}}{t^{6} + t^{4} - 5 t^{2} + 3} d t =$ $= 32 \int \frac{t^{2}}{{(t - 1)}^{2} {(t + 1)}^{2} (t^{2} + 3)} d t =$ $= 2 \int \frac{d t}{{(t - 1)}^{2}} + 2 \int \frac{d t}{{(t + 1)}^{2}} - 6 \int \frac{d t}{t^{2} + 3} + \int \frac{d t}{t - 1} - \int \frac{d t}{t + 1} =$ $= - \frac{2}{t - 1} - \frac{2}{t + 1} - 2 \sqrt{3} \arctan \frac{\sqrt{3} t}{3} + \ln (t - 1) - \ln (t + 1) =$ $= - \frac{4 t}{t^{2} - 1} - 2 \sqrt{3} \arctan \frac{\sqrt{3} t}{3} + \ln \frac{t - 1}{t + 1} =$ $= \frac{\sqrt{(x^{2} - x + 1) (x^{2} + 3 x + 1)}}{x} - 2 \sqrt{3} \arctan \frac{\sqrt{3 (x^{2} - x + 1)}}{3 \sqrt{x^{2} + 3 x + 1}} + \ln ∣ \frac{x^{2} + x + 1 - \sqrt{(x^{2} - x + 1) (x^{2} + 3 x + 1)}}{2 x} ∣ + C$
	Commented by liberty last updated on 29/Nov/20

	$w a w g r e a t$
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