find ∫ (x^2 +1)(√((x+1)/(x−2)))dx


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Question Number 63822 by mathmax by abdo last updated on 09/Jul/19

$f i n d \int (x^{2} + 1) \sqrt{\frac{x + 1}{x - 2}} d x$
Commented by mathmax by abdo last updated on 11/Jul/19
$let A =∫ (x^2 +1)(√((x+1)/(x−2)))dx changement (√((x+1)/(x−2)))=t give ((x+1)/(x−2)) =t^(2 ) ⇒((x−2 +3)/(x−2)) =t^2 ⇒1+(3/(x−2)) =t^2 ⇒(3/(x−2)) =t^2 −1 ⇒ ((x−2)/3) =(1/(t^2 −1)) ⇒x−2 =(3/(t^2 −1)) ⇒dx =((−6t)/((t^2 −1)^2 ))dt ⇒ A = ∫ {((3/(t^2 −1)) +2)^2 +1}t ((−6t)/((t^2 −1)^2 ))dt =∫ ((−6t^2 )/((t^2 −1)^2 )){ (((2t^2 +1)/(t^2 −1)))^(2 ) +1}dt =∫ ((−6t^2 )/((t^2 −1)^2 ))(((2t^2 +1)^2 +(t^2 −1)^2 )/((t^2 −1)^2 ))dt =∫ ((−6t^2 { 4t^4 +4t^2 +1+t^4 −2t^2 +1})/((t^2 −1)^4 ))dt =−6 ∫ ((t^2 {5t^4 +2t^2 +2})/((t^2 −1)^4 )) dt =−6 ∫ ((5t^6 +2t^4 +2t^2 )/((t^2 −1)^4 ))dt let decompose F(t)=((5t^6 +2t^4 +2t^2 )/((t^2 −1)^4 )) =(((...))/((t−1)^4 (t+1)^4 )) =Σ_(i=1) ^4 (a_i /((t−1)^i )) +Σ_(i=1) ^4 (b_i /((t+1)^i )) changement t−1 =u give F(t)=G(u) =((5(u+1)^6 +2(u+1)^4 +2(u+1)^2 )/(u^4 (u+2)^4 )) let find D_3 (0) for f(u) =(1/((u+2)^4 )) =(u+2)^(−4) ⇒f(u) =f(0)+uf^′ (0) +(u^2 /2)f^((2)) (0) +(u^3 /(3!))f^((3)) (0) +(u^3 /(3!))ξ(u) f(0) =2^(−4) ,f^′ (u)=−4(u+2)^(−5) ⇒f^′ (0) =−4.2^(−5) f^((2)) (u) =20(u+2)^(−6) ⇒f^((2)) (0) =20.2^(−6) f^((3)) (u) =−120(u+2)^(−7) ⇒f^((3)) (0) =−120 .2^(−7) ⇒ f(u) =2^(−4) −4.2^(−5) u +10. 2^(−6) u^2 −20. 2^(−7) u^3 +(u^3 /(3!))ξ(u) ⇒ G(u) =((5(u+1)^6 +2(u+1)^4 +2(u+1)^2 )/u^4 ){2^(−4) −4.2^(−5) u +10.2^(−6) u^2 −20.2^(−7) u^3 +(u^3 /(3!))ξ(u)}.....be continued....$
$l e t A = \int (x^{2} + 1) \sqrt{\frac{x + 1}{x - 2}} d x c h a n g e m e n t \sqrt{\frac{x + 1}{x - 2}} = t g i v e$ $\frac{x + 1}{x - 2} = t^{2} \Rightarrow \frac{x - 2 + 3}{x - 2} = t^{2} \Rightarrow 1 + \frac{3}{x - 2} = t^{2} \Rightarrow \frac{3}{x - 2} = t^{2} - 1 \Rightarrow$ $\frac{x - 2}{3} = \frac{1}{t^{2} - 1} \Rightarrow x - 2 = \frac{3}{t^{2} - 1} \Rightarrow d x = \frac{- 6 t}{{(t^{2} - 1)}^{2}} d t \Rightarrow$ $A = \int {{(\frac{3}{t^{2} - 1} + 2)}^{2} + 1} t \frac{- 6 t}{{(t^{2} - 1)}^{2}} d t$ $= \int \frac{- 6 t^{2}}{{(t^{2} - 1)}^{2}} {{(\frac{2 t^{2} + 1}{t^{2} - 1})}^{2} + 1} d t$ $= \int \frac{- 6 t^{2}}{{(t^{2} - 1)}^{2}} \frac{{(2 t^{2} + 1)}^{2} + {(t^{2} - 1)}^{2}}{{(t^{2} - 1)}^{2}} d t$ $= \int \frac{- 6 t^{2} {4 t^{4} + 4 t^{2} + 1 + t^{4} - 2 t^{2} + 1}}{{(t^{2} - 1)}^{4}} d t$ $= - 6 \int \frac{t^{2} {5 t^{4} + 2 t^{2} + 2}}{{(t^{2} - 1)}^{4}} d t = - 6 \int \frac{5 t^{6} + 2 t^{4} + 2 t^{2}}{{(t^{2} - 1)}^{4}} d t$ $l e t d e c o m p o s e F (t) = \frac{5 t^{6} + 2 t^{4} + 2 t^{2}}{{(t^{2} - 1)}^{4}} = \frac{(. . .)}{{(t - 1)}^{4} {(t + 1)}^{4}}$ $= \sum_{i = 1}^{4} \frac{a_{i}}{{(t - 1)}^{i}} + \sum_{i = 1}^{4} \frac{b_{i}}{{(t + 1)}^{i}} c h a n g e m e n t t - 1 = u g i v e$ $F (t) = G (u) = \frac{5 {(u + 1)}^{6} + 2 {(u + 1)}^{4} + 2 {(u + 1)}^{2}}{u^{4} {(u + 2)}^{4}} l e t f i n d D_{3} (0) f o r$ $f (u) = \frac{1}{{(u + 2)}^{4}} = {(u + 2)}^{- 4} \Rightarrow f (u) = f (0) + {u f}^{^{'}} (0) + \frac{u^{2}}{2} f^{(2)} (0)$ $+ \frac{u^{3}}{3!} f^{(3)} (0) + \frac{u^{3}}{3!} ξ (u)$ $f (0) = 2^{- 4}, f^{^{'}} (u) = - 4 {(u + 2)}^{- 5} \Rightarrow f^{^{'}} (0) = - 4 . 2^{- 5}$ $f^{(2)} (u) = 20 {(u + 2)}^{- 6} \Rightarrow f^{(2)} (0) = 20 . 2^{- 6}$ $f^{(3)} (u) = - 120 {(u + 2)}^{- 7} \Rightarrow f^{(3)} (0) = - 120 . 2^{- 7} \Rightarrow$ $f (u) = 2^{- 4} - 4 . 2^{- 5} u + 10 . 2^{- 6} u^{2} - 20 . 2^{- 7} u^{3} + \frac{u^{3}}{3!} ξ (u) \Rightarrow$ $G (u) = \frac{5 {(u + 1)}^{6} + 2 {(u + 1)}^{4} + 2 {(u + 1)}^{2}}{u^{4}} {2^{- 4} - 4 . 2^{- 5} u + 10 . 2^{- 6} u^{2}$ $- 20 . 2^{- 7} u^{3} + \frac{u^{3}}{3!} ξ (u)} . . . . . b e c o n t i n u e d . . . .$
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