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Question Number 46156 by Meritguide1234 last updated on 21/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18
	$x^4 +2x^3 −x^2 +2x+1 =x^2 (x^2 +(1/x^2 )+2x+(2/x)−1) =x^2 {(x+(1/x))^2 −2+2(x+(1/x))−1} =x^2 {(x+(1/x))^2 +2(x+(1/x))−3} ∫(((1−(1/x^2 ))×x(√({(x+(1/x))^2 +2(x+(1/x))−3)) )/((x+1)^2 )) dx ∫(((1−(1/x^2 ))×x×(√({(x+(1/x))^2 +2(x+(1/x))−3)) dx)/(x(x+2+(1/x)))) t=x+(1/x) dt=1−(1/x^2 )dx ∫((√(t^2 +2t−3))/(t+2))dt ←standard form ∫((t^2 +2t−3)/((t+2)(√(t^2 +2t−3)) ))dt ∫((tdt)/(√(t^2 +2t−3)))−∫((3dt)/((t+2)(√(t^2 +2t−3)))) (1/2)∫((2t+2−2)/(√(t^2 +2t−3)))−3∫(dt/((t+2)(√(t^2 +2t−3)) )) (1/2)∫((d(t^2 +2t−3))/(√(t^2 +2t−3)))−∫(dt/(√((t+1)^2 −4)))−∫(dt/((t+2)(√(t^2 +2t−3)))) (1/2)×(((t^2 +2t−3)^(((−1)/2)+1) )/(1/2))−ln{(t+1)+(√(t^2 +2t−1)) +I_3 ={(x+(1/x))^2 +2(x+(1/x))−3}^(1/2) −ln{(x+(1/x)+1)+(√((x+(1/x))^2 +2(x+(1/x))1)) +I_3 now calcukating I_3 ∫(dt/((t+2)(√(t^2 +2t−3)))) t+2=(1/k) dt=((−1)/k^2 )dk ∫((−dk)/(k^2 ×(1/k)(√(((1/k)−2)^2 +2((1/k)−2)−3)))) ∫((−dk)/(k(√((1/k^2 )−(4/k)+4+(2/k)−4−3)))) ∫((−dk)/(k(√((1−4k+4k^2 +2k−7k^2 )/k^2 )))) ∫((−dk)/(√(−3k^2 −2k+1))) ∫((−dk)/(√(1−3(k^2 +(2/3)k+(1/9)−(1/9))))) ∫((−dk)/(√(1−3{(k+(1/3))^2 −(1/9)}))) ∫((−dk)/(√(1+(1/3)−3(k+(1/3))^2 ))) ∫((−dk)/(√((4/3)−3(k+(1/3))^2 ))) (1/(√3))∫((−dk)/(√(((2/3))^2 −(k+(1/3))^2 ))) =((−1)/(√3))×sin^(−1) (((k+(1/3))/(2/3))) =((−1)/(√3))sin^(−1) ((((1/(t+2))+(1/3))/(2/3))) ((−1)/(√3))sin^(−1) ((((1/(x+(1/x)+2))+(1/3))/(2/3)))+c$
	$x^{4} + 2 x^{3} - x^{2} + 2 x + 1$ $= x^{2} (x^{2} + \frac{1}{x^{2}} + 2 x + \frac{2}{x} - 1)$ $= x^{2} {{(x + \frac{1}{x})}^{2} - 2 + 2 (x + \frac{1}{x}) - 1}$ $= x^{2} {{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3}$ $\int \frac{(1 - \frac{1}{x^{2}}) \times x \sqrt{{{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3}}{{(x + 1)}^{2}} d x$ $\int \frac{(1 - \frac{1}{x^{2}}) \times x \times \sqrt{{{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3} d x}{x (x + 2 + \frac{1}{x})}$ $t = x + \frac{1}{x} d t = 1 - \frac{1}{x^{2}} d x$ $\int \frac{\sqrt{t^{2} + 2 t - 3}}{t + 2} d t \leftarrow s t a n d a r d f o r m$ $\int \frac{t^{2} + 2 t - 3}{(t + 2) \sqrt{t^{2} + 2 t - 3}} d t$ $\int \frac{t d t}{\sqrt{t^{2} + 2 t - 3}} - \int \frac{3 d t}{(t + 2) \sqrt{t^{2} + 2 t - 3}}$ $\frac{1}{2} \int \frac{2 t + 2 - 2}{\sqrt{t^{2} + 2 t - 3}} - 3 \int \frac{d t}{(t + 2) \sqrt{t^{2} + 2 t - 3}}$ $\frac{1}{2} \int \frac{d (t^{2} + 2 t - 3)}{\sqrt{t^{2} + 2 t - 3}} - \int \frac{d t}{\sqrt{{(t + 1)}^{2} - 4}} - \int \frac{d t}{(t + 2) \sqrt{t^{2} + 2 t - 3}}$ $\frac{1}{2} \times \frac{{(t^{2} + 2 t - 3)}^{\frac{- 1}{2} + 1}}{\frac{1}{2}} - l n {(t + 1) + \sqrt{t^{2} + 2 t - 1} + I_{3}$ $= {{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) - 3}^{\frac{1}{2}} - l n {(x + \frac{1}{x} + 1) + \sqrt{{(x + \frac{1}{x})}^{2} + 2 (x + \frac{1}{x}) 1} + I_{3}$ $n o w c a l c u k a t i n g I_{3}$ $\int \frac{d t}{(t + 2) \sqrt{t^{2} + 2 t - 3}}$ $t + 2 = \frac{1}{k} d t = \frac{- 1}{k^{2}} d k$ $\int \frac{- d k}{k^{2} \times \frac{1}{k} \sqrt{{(\frac{1}{k} - 2)}^{2} + 2 (\frac{1}{k} - 2) - 3}}$ $\int \frac{- d k}{k \sqrt{\frac{1}{k^{2}} - \frac{4}{k} + 4 + \frac{2}{k} - 4 - 3}}$ $\int \frac{- d k}{k \sqrt{\frac{1 - 4 k + 4 k^{2} + 2 k - 7 k^{2}}{k^{2}}}}$ $\int \frac{- d k}{\sqrt{- 3 k^{2} - 2 k + 1}}$ $\int \frac{- d k}{\sqrt{1 - 3 (k^{2} + \frac{2}{3} k + \frac{1}{9} - \frac{1}{9})}}$ $\int \frac{- d k}{\sqrt{1 - 3 {{(k + \frac{1}{3})}^{2} - \frac{1}{9}}}}$ $\int \frac{- d k}{\sqrt{1 + \frac{1}{3} - 3 {(k + \frac{1}{3})}^{2}}}$ $\int \frac{- d k}{\sqrt{\frac{4}{3} - 3 {(k + \frac{1}{3})}^{2}}}$ $\frac{1}{\sqrt{3}} \int \frac{- d k}{\sqrt{{(\frac{2}{3})}^{2} - {(k + \frac{1}{3})}^{2}}}$ $= \frac{- 1}{\sqrt{3}} \times {s i n}^{- 1} (\frac{k + \frac{1}{3}}{\frac{2}{3}})$ $= \frac{- 1}{\sqrt{3}} {s i n}^{- 1} (\frac{\frac{1}{t + 2} + \frac{1}{3}}{\frac{2}{3}})$ $\frac{- 1}{\sqrt{3}} {s i n}^{- 1} (\frac{\frac{1}{x + \frac{1}{x} + 2} + \frac{1}{3}}{\frac{2}{3}}) + c$
	Commented by Meritguide1234 last updated on 22/Oct/18

	$v e r y n i c e$
	Commented by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18

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