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Question Number 60212 by peter frank last updated on 18/May/19

Commented by maxmathsup by imad last updated on 19/May/19
$we have 3x^3 −x^2 +2x−4 =3x^3 −3x^2 +2x^2 +2x−4 =3x^2 (x−1) +2x^2 −2x +4x−4 =3x^2 (x−1) +2x(x−1)+4(x−1) =(x−1)(3x^2 +2x +4) also x^2 −3x +2 =x^2 −x −2x+2 =x(x−1)−2(x−1) =(x−1)(x−2) ⇒ I =∫_0 ^1 (((x−1)(3x^2 +2x+4))/((√(1−x))(√(2−x)))) dx =− ∫_0 ^1 (((√(1−x))(3x^2 +2x+4))/(√(2−x))) dx =−∫_0 ^1 (√((1−x)/(2−x)))(3x^2 +2x +4)dx changement (√((1−x)/(2−x)))=t give ((1−x)/(2−x)) =t^2 ⇒1−x =2t^2 −t^2 x ⇒1−2t^2 =(1−t^2 )x ⇒x =((1−2t^2 )/(1−t^2 )) ⇒ dx/dt =((−4t(1−t^2 )−(1−2t^2 )(−2t))/((1−t^2 )^2 )) =((−4t+4t^3 +2t−4t^3 )/((1−t^2 )^2 )) =((−2t)/((1−t^2 )^2 )) ⇒ I =−∫_(1/(√2)) ^0 t{ 3(((1−2t^2 )/(1−t^2 )))^2 +2((1−2t^2 )/(1−t^2 )) +4}(((−2t)/((1−t^2 )^2 )))dt I =∫_0 ^(1/(√2)) −2t^2 { ((3(1−2t^2 )^2 +2(1−2t^2 )(1−t^2 ) +4(1−t^2 )^2 )/((1−t^2 )^4 ))}dt I =−2 ∫_0 ^(1/(√2)) (t^2 /((1−t^2 )^4 )){ 3(4t^4 −4t^2 +1)+2(1−t^2 −2t^2 +2t^4 ) +4(t^4 −2t^2 +1)}dt =−2 ∫_0 ^(1/(√2)) (t^2 /((1−t^2 )^4 )){12t^4 −12t^2 +3 +2 −6t^2 +4t^4 +4t^4 −8t^2 +4}dt =−2 ∫_0 ^(1/(√2)) (t^2 /((1−t^2 )^4 )){ 20t^4 −26t^2 +9} dt =−2 ∫_0 ^(1/(√2)) ((20t^6 −26t^4 +9t^2 )/((1−t^2 )^4 )) dt let decompose F(t) =((20t^6 −26 t^4 +9t^2 )/((t^2 −1)^4 )) ⇒F(t) =(a/(t−1)) +(b/((t−1)^2 )) +(c/((t−1)^3 )) +(d/((t−1)^4 )) +(e/(t+1)) +(f/((t+1)^2 )) +(g/((t+1)^3 )) +(h/((t+1)^4 )) F(−t)=F(t) ⇒((−a)/(t+1)) +(b/((t+1)^2 )) −(c/((t+1)^3 )) +(d/((t+1)^4 )) −(e/(t−1)) +(f/((t−1)^2 )) −(g/((t−1)^3 )) +(h/((t−1)^4 )) =F(t) ⇒−a =e ,b=f , −c=g ,d =h ,....be continued...$
$w e h a v e 3 x^{3} - x^{2} + 2 x - 4 = 3 x^{3} - 3 x^{2} + 2 x^{2} + 2 x - 4$ $= 3 x^{2} (x - 1) + 2 x^{2} - 2 x + 4 x - 4 = 3 x^{2} (x - 1) + 2 x (x - 1) + 4 (x - 1)$ $= (x - 1) (3 x^{2} + 2 x + 4) a l s o$ $x^{2} - 3 x + 2 = x^{2} - x - 2 x + 2 = x (x - 1) - 2 (x - 1) = (x - 1) (x - 2) \Rightarrow$ $I = \int_{0}^{1} \frac{(x - 1) (3 x^{2} + 2 x + 4)}{\sqrt{1 - x} \sqrt{2 - x}} d x = - \int_{0}^{1} \frac{\sqrt{1 - x} (3 x^{2} + 2 x + 4)}{\sqrt{2 - x}} d x$ $= - \int_{0}^{1} \sqrt{\frac{1 - x}{2 - x}} (3 x^{2} + 2 x + 4) d x c h a n g e m e n t \sqrt{\frac{1 - x}{2 - x}} = t g i v e$ $\frac{1 - x}{2 - x} = t^{2} \Rightarrow 1 - x = 2 t^{2} - t^{2} x \Rightarrow 1 - 2 t^{2} = (1 - t^{2}) x \Rightarrow x = \frac{1 - 2 t^{2}}{1 - t^{2}} \Rightarrow$ $d x / d t = \frac{- 4 t (1 - t^{2}) - (1 - 2 t^{2}) (- 2 t)}{{(1 - t^{2})}^{2}} = \frac{- 4 t + 4 t^{3} + 2 t - 4 t^{3}}{{(1 - t^{2})}^{2}} = \frac{- 2 t}{{(1 - t^{2})}^{2}} \Rightarrow$ $I = - \int_{\frac{1}{\sqrt{2}}}^{0} t {3 {(\frac{1 - 2 t^{2}}{1 - t^{2}})}^{2} + 2 \frac{1 - 2 t^{2}}{1 - t^{2}} + 4} (\frac{- 2 t}{{(1 - t^{2})}^{2}}) d t$ $I = \int_{0}^{\frac{1}{\sqrt{2}}} - 2 t^{2} {\frac{3 {(1 - 2 t^{2})}^{2} + 2 (1 - 2 t^{2}) (1 - t^{2}) + 4 {(1 - t^{2})}^{2}}{{(1 - t^{2})}^{4}}} d t$ $I = - 2 \int_{0}^{\frac{1}{\sqrt{2}}} \frac{t^{2}}{{(1 - t^{2})}^{4}} {3 (4 t^{4} - 4 t^{2} + 1) + 2 (1 - t^{2} - 2 t^{2} + 2 t^{4}) + 4 (t^{4} - 2 t^{2} + 1)} d t$ $= - 2 \int_{0}^{\frac{1}{\sqrt{2}}} \frac{t^{2}}{{(1 - t^{2})}^{4}} {12 t^{4} - 12 t^{2} + 3 + 2 - 6 t^{2} + 4 t^{4} + 4 t^{4} - 8 t^{2} + 4} d t$ $= - 2 \int_{0}^{\frac{1}{\sqrt{2}}} \frac{t^{2}}{{(1 - t^{2})}^{4}} {20 t^{4} - 26 t^{2} + 9} d t$ $= - 2 \int_{0}^{\frac{1}{\sqrt{2}}} \frac{20 t^{6} - 26 t^{4} + 9 t^{2}}{{(1 - t^{2})}^{4}} d t l e t d e c o m p o s e$ $F (t) = \frac{20 t^{6} - 26 t^{4} + 9 t^{2}}{{(t^{2} - 1)}^{4}} \Rightarrow F (t) = \frac{a}{t - 1} + \frac{b}{{(t - 1)}^{2}} + \frac{c}{{(t - 1)}^{3}} + \frac{d}{{(t - 1)}^{4}} + \frac{e}{t + 1}$ $+ \frac{f}{{(t + 1)}^{2}} + \frac{g}{{(t + 1)}^{3}} + \frac{h}{{(t + 1)}^{4}}$ $F (- t) = F (t) \Rightarrow \frac{- a}{t + 1} + \frac{b}{{(t + 1)}^{2}} - \frac{c}{{(t + 1)}^{3}} + \frac{d}{{(t + 1)}^{4}} - \frac{e}{t - 1} + \frac{f}{{(t - 1)}^{2}} - \frac{g}{{(t - 1)}^{3}}$ $+ \frac{h}{{(t - 1)}^{4}} = F (t) \Rightarrow - a = e, b = f, - c = g, d = h, . . . . b e c o n t i n u e d . . .$
	Answered by MJS last updated on 19/May/19

	$\int \frac{3 x^{3} - x^{2} + 2 x - 4}{\sqrt{x^{2} - 3 x + 2}} d x =$ $[t = \sqrt{x^{2} - 3 x + 2} \to d x = \frac{2 t d t}{\sqrt{4 t^{2} + 1}}]$ $= \int (\frac{25 t^{2}}{\sqrt{4 t^{2} + 1}} + \frac{20}{\sqrt{4 t^{2} + 1}} + 3 t^{2} + 20) d t =$ $= \int [(\frac{25 (8 t^{2} + 1)}{8 \sqrt{4 t^{2} + 1}} - \frac{25}{8 \sqrt{4 t^{2} + 1}}) + \frac{20}{\sqrt{4 t^{2} + 1}} + 3 t^{2} + 20] d t =$ $= \int (\frac{25 (8 t^{2} + 1)}{8 \sqrt{4 t^{2} + 1}} + \frac{135}{8 \sqrt{4 t^{2} + 1}} + 3 t^{2} + 20) d t$ $\frac{25}{8} \int \frac{8 t^{2} + 1}{\sqrt{4 t^{2} + 1}} d t =$ $[u = arcsinh 2 t \to d t = \frac{\sqrt{4 t^{2} + 1}}{2} d u]$ $= \frac{25}{16} \int \cosh 2 u d u = \frac{25}{32} \sinh 2 u = \frac{25}{8} t \sqrt{4 t^{2} + 1} =$ $= \frac{25}{8} (2 x - 3) \sqrt{x^{2} - 3 x + 2}$ $\frac{135}{8} \int \frac{d t}{\sqrt{4 t^{2} + 1}} =$ $[u = arcsinh 2 t \to d t = \frac{\sqrt{4 t^{2} + 1}}{2} d u]$ $= \frac{135}{16} \int d u = \frac{135}{16} u = \frac{135}{16} arcsinh 2 t =$ $= \frac{135}{16} arcsinh 2 \sqrt{x^{2} - 3 x + 2}$ $\int (3 t^{2} + 20) d t = t^{3} + 20 t = (x^{3} - 3 x + 22) \sqrt{x^{2} - 3 x + 2}$ $\int \frac{3 x^{3} - x^{2} + 2 x - 4}{\sqrt{x^{2} - 3 x + 2}} d x =$ $= \frac{1}{8} (8 x^{2} + 26 x + 101) \sqrt{x^{2} - 3 x + 2} + \frac{135}{16} arcsinh 2 \sqrt{x^{2} - 3 x + 2} + C$
	Commented by peter frank last updated on 19/May/19

	$t h a n k s . s o r r y s i r e x p l a i n$ $t h i r d l i n e$
	Commented by MJS last updated on 19/May/19

	$I finished it$
	Commented by maxmathsup by imad last updated on 19/May/19

	$t h a n k y o u s i r f o r t h i s h a r d w o r k .$
	Commented by peter frank last updated on 24/May/19

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