calculate ∫_2 ^5 ((xdx)/(2x+1 +(√(x−1))))


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Question Number 35990 by abdo mathsup 649 cc last updated on 26/May/18

$c a l c u l a t e \int_{2}^{5} \frac{x d x}{2 x + 1 + \sqrt{x - 1}}$
Commented by abdo mathsup 649 cc last updated on 27/May/18
$let put I = ∫_2 ^5 ((xdx)/(2x+1+(√(x−1)))) changement (√(x−1))=t give x =t^2 +1 and I = ∫_1 ^2 ((t^2 +1)/(2t^2 +2+1 +t)) (2t)dt = ∫_1 ^2 ((2t^3 +2t)/(2t^2 +t+3))dt = ∫_1 ^2 ((t( 2t^2 +t+3) −t^2 −3t +2t)/(2t^2 +t +3))dt = ∫_1 ^2 t dt −∫_1 ^2 ((t^2 +t)/(2t^2 +t +3))dt but ∫_1 ^2 t dt =[(t^2 /2)]_1 ^2 = 2−(1/2) =(3/2) ∫_1 ^2 ((t^2 +t)/(2t^2 +t +3))dt =(1/2) ∫_1 ^2 ((2t^2 +t+t +3−3)/(2t^2 +t +3))dt =(1/2) ∫_1 ^2 dt +(1/2) ∫_1 ^2 ((t−3)/(2t^2 +t+3))dt =(1/2) +(1/8)∫_1 ^2 ((4t −12)/(2t^2 +t +3))dt =(1/2) +(1/8) J J = ∫_1 ^2 ((4t+1−13)/(2t^2 +t+3))dt =[ln(2t^2 +t+3)]_1 ^2 −13 ∫_1 ^2 (dt/(2t^2 +t+3)) = ln(13)−ln(6) −((13)/2) ∫_1 ^2 (dt/(t^2 +(t/2) +(3/2))) but t^2 +(t/2) +(3/2) =t^2 +2(1/4)t +(1/(16)) +(3/2) −(1/(16)) =(t +(1/4))^2 + ((23)/(16)) chang. t+(1/4) =((√(23))/4) x give ∫_1 ^2 (dt/(t^2 +(t/2)+(3/2))) = ∫_((1/(√(23)))(5)) ^(9/(√(23))) (1/(((23)/(16))(1+x^2 ))) ((√(23))/4) dx = ((16)/(23)) ((√(23))/4) ∫_(5/(√(23))) ^(9/(√(23))) (dx/(1+x^2 )) =(4/(√(23))){ arctan((9/(√(2)))))−arctan((5/(√(23))))} I =(3/2) −(1/2) −(1/8)J =1 −(1/8){ln(13)−ln(6)−((13)/2)(4/(√(23))){arctan((9/(√(23))))−arctan((5/(√(23))))}o$
$l e t p u t I = \int_{2}^{5} \frac{x d x}{2 x + 1 + \sqrt{x - 1}} c h a n g e m e n t$ $\sqrt{x - 1} = t g i v e x = t^{2} + 1 a n d$ $I = \int_{1}^{2} \frac{t^{2} + 1}{2 t^{2} + 2 + 1 + t} (2 t) d t$ $= \int_{1}^{2} \frac{2 t^{3} + 2 t}{2 t^{2} + t + 3} d t$ $= \int_{1}^{2} \frac{t (2 t^{2} + t + 3) - t^{2} - 3 t + 2 t}{2 t^{2} + t + 3} d t$ $= \int_{1}^{2} t d t - \int_{1}^{2} \frac{t^{2} + t}{2 t^{2} + t + 3} d t b u t$ $\int_{1}^{2} t d t = {[\frac{t^{2}}{2}]}_{1}^{2} = 2 - \frac{1}{2} = \frac{3}{2}$ $\int_{1}^{2} \frac{t^{2} + t}{2 t^{2} + t + 3} d t = \frac{1}{2} \int_{1}^{2} \frac{2 t^{2} + t + t + 3 - 3}{2 t^{2} + t + 3} d t$ $= \frac{1}{2} \int_{1}^{2} d t + \frac{1}{2} \int_{1}^{2} \frac{t - 3}{2 t^{2} + t + 3} d t$ $= \frac{1}{2} + \frac{1}{8} \int_{1}^{2} \frac{4 t - 12}{2 t^{2} + t + 3} d t = \frac{1}{2} + \frac{1}{8} J$ $J = \int_{1}^{2} \frac{4 t + 1 - 13}{2 t^{2} + t + 3} d t = {[l n (2 t^{2} + t + 3)]}_{1}^{2}$ $- 13 \int_{1}^{2} \frac{d t}{2 t^{2} + t + 3} = l n (13) - l n (6)$ $- \frac{13}{2} \int_{1}^{2} \frac{d t}{t^{2} + \frac{t}{2} + \frac{3}{2}} b u t$ $t^{2} + \frac{t}{2} + \frac{3}{2} = t^{2} + 2 \frac{1}{4} t + \frac{1}{16} + \frac{3}{2} - \frac{1}{16}$ $= {(t + \frac{1}{4})}^{2} + \frac{23}{16} c h a n g . t + \frac{1}{4} = \frac{\sqrt{23}}{4} x g i v e$ $\int_{1}^{2} \frac{d t}{t^{2} + \frac{t}{2} + \frac{3}{2}} = \int_{\frac{1}{\sqrt{23}} (5)}^{\frac{9}{\sqrt{23}}} \frac{1}{\frac{23}{16} (1 + x^{2})} \frac{\sqrt{23}}{4} d x$ $= \frac{16}{23} \frac{\sqrt{23}}{4} \int_{\frac{5}{\sqrt{23}}}^{\frac{9}{\sqrt{23}}} \frac{d x}{1 + x^{2}} = \frac{4}{\sqrt{23}} {a r c t a n (\frac{9}{\sqrt{2)}}) - a r c t a n (\frac{5}{\sqrt{23}})}$ $I = \frac{3}{2} - \frac{1}{2} - \frac{1}{8} J$ $= 1 - \frac{1}{8} {l n (13) - l n (6) - \frac{13}{2} \frac{4}{\sqrt{23}} {a r c t a n (\frac{9}{\sqrt{23}}) - a r c t a n (\frac{5}{\sqrt{23}})} o$
Commented by abdo mathsup 649 cc last updated on 27/May/18

$I = 1 - \frac{l n (13)}{8} + \frac{l n (6)}{8} + \frac{13}{4 \sqrt{23}} {a r c t a n (\frac{9}{\sqrt{23}}) - a r c t a n (\frac{5}{\sqrt{23}})} .$
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