find ∫ (√(x+2−(√(x−1))))dx


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Question Number 47018 by maxmathsup by imad last updated on 03/Nov/18

$f i n d \int \sqrt{x + 2 - \sqrt{x - 1}} d x$
Commented by maxmathsup by imad last updated on 04/Nov/18

$l e t A = \int \sqrt{x + 2 - \sqrt{x - 1}} d x c h a n g e m e n t \sqrt{x - 1} = t g i v e$ $A = \int \sqrt{1 + t^{2} + 2 - t} (2 t) d t = 2 \int t \sqrt{t^{2} - t + 3} d t b u t$ $t^{2} - t + 3 = {(t - \frac{1}{2})}^{2} + 3 - \frac{1}{4} = {(t - \frac{1}{2})}^{2} + \frac{11}{4} \Rightarrow A =_{t - \frac{1}{2} = \frac{\sqrt{11}}{2} u} \int (\frac{1}{2} + \frac{\sqrt{11}}{2} u) \frac{\sqrt{11}}{2} \sqrt{1 + u^{2}} \frac{\sqrt{11}}{2} d u$ $= \frac{11}{4} \int (1 + \sqrt{11} u) \sqrt{1 + u^{2}} d u \Rightarrow \frac{4}{11} A =_{u = s h α} \int (1 + \sqrt{11} s h α) c h α c h α d α$ $= \int {c h}^{2} α d α + \sqrt{11} \int s h α {c h}^{2} α d α$ $= \int \frac{1 + c h (2 α)}{2} d α + \frac{\sqrt{11}}{3} {c h}^{3} α + C = \frac{α}{2} + \frac{1}{4} s h (α) + \frac{\sqrt{11}}{3} {c h}^{3} α + C$ $= \frac{a r g s h (u)}{2} + \frac{u}{4} + \frac{\sqrt{11}}{3} {c h}^{3} (a r g s h (u)) + C$ $= \frac{1}{2} l n (u + \sqrt{1 + u^{2}}) + \frac{u}{4} + \frac{\sqrt{11}}{3} {c h}^{3} (l n (u + \sqrt{1 + u^{2}}) + C$ $= \frac{1}{2} l n (\frac{2 t - 1}{\sqrt{11}} + \sqrt{1 + {(\frac{2 t - 1}{\sqrt{11}})}^{2}}) + \frac{2 t - 1}{4 \sqrt{11}} + \frac{\sqrt{11}}{3} {c h}^{3} (l n (\frac{2 t - 1}{\sqrt{11}} + \sqrt{1 + (\frac{2 t - 1}{\sqrt{11}})}) + C$ $w i t h t = \sqrt{x - 1} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18
	$x−1=t^2 dx=2tdt ∫(√(t^2 +1+2−t)) ×2tdt ∫(2t−1+1)(√(t^2 −t+3)) dt ∫(√(t^2 −t+3)) d(t^2 −t+1)+∫(√(t^2 −2×t×(1/2)+(1/4)+3−(1/4) )) dt I_1 +I_2 I_1 =(((t^2 −t+1)^(3/2) )/(3/2))+c_1 I_1 =(((x−(√(x−1)) )^(3/2) )/(3/2))+c_1 I_2 =∫(√((t−(1/2))^2 +(((√(11))/2))^2 )) dt =(((t−(1/2)))/2)×(√((t−(1/2))^2 +(((√(11))/2))^2 )) +(((((√(11))/2))^2 )/2)ln{(t−(1/2))+(√((t−(1/2))^2 +(((√(11))/2))^2 )) +c_2 =((((√(x−1)) −(1/2)))/2)×(√(((√(x−1)) −(1/2))^2 +((((√(11)) )/2))^2 )) +(((((√(11))/2))^2 )/2)ln{((√(x−1)) −(1/2))+(√(((√(x−1)) −(1/2))^2 +(((√(11))/2))^2 )) +c_2$
	$x - 1 = t^{2}$ $d x = 2 t d t$ $\int \sqrt{t^{2} + 1 + 2 - t} \times 2 t d t$ $\int (2 t - 1 + 1) \sqrt{t^{2} - t + 3} d t$ $\int \sqrt{t^{2} - t + 3} d (t^{2} - t + 1) + \int \sqrt{t^{2} - 2 \times t \times \frac{1}{2} + \frac{1}{4} + 3 - \frac{1}{4}} d t$ $I_{1} + I_{2}$ $I_{1} = \frac{{(t^{2} - t + 1)}^{\frac{3}{2}}}{\frac{3}{2}} + c_{1}$ $I_{1} = \frac{{(x - \sqrt{x - 1})}^{\frac{3}{2}}}{\frac{3}{2}} + c_{1}$ $I_{2} = \int \sqrt{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}} d t$ $= \frac{(t - \frac{1}{2})}{2} \times \sqrt{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}} + \frac{{(\frac{\sqrt{11}}{2})}^{2}}{2} l n {(t - \frac{1}{2}) + \sqrt{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}} + c_{2}$ $= \frac{(\sqrt{x - 1} - \frac{1}{2})}{2} \times \sqrt{{(\sqrt{x - 1} - \frac{1}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}} + \frac{{(\frac{\sqrt{11}}{2})}^{2}}{2} l n {(\sqrt{x - 1} - \frac{1}{2}) + \sqrt{{(\sqrt{x - 1} - \frac{1}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}} + c_{2}$
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