find I = ∫_1 ^2 (dx/(x(√(x+1)) +(x+1)(√x)))


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Question Number 36397 by prof Abdo imad last updated on 01/Jun/18

$f i n d I = \int_{1}^{2} \frac{d x}{x \sqrt{x + 1} + (x + 1) \sqrt{x}}$
	Answered by behi83417@gmail.com last updated on 01/Jun/18

	$x = {t g}^{2} t \Rightarrow d x = 2 t g t (1 + {t g}^{2} t) d t$ $\frac{1}{{c o s}^{2} t} = 1 + x \Rightarrow c o s t = \frac{1}{\sqrt{1 + x}}$ $I = \int \frac{2 t g t (1 + {t g}^{2} t) d t}{{t g}^{2} t . \frac{1}{c o s t} + \frac{1}{{c o s}^{2} t} . t g t} = \int \frac{2 t g t (1 + {t g}^{2} t) d t}{\frac{{s i n}^{2} t + s i n t}{{c o s}^{3} t}} =$ $= \int \frac{2 s i n t d t}{{s i n}^{2} t + s i n t} = \int \frac{2 d t}{1 + s i n t} = \frac{- 4}{1 + t g \frac{t}{2}} + c$ $t g \frac{t}{2} = \sqrt{\frac{1 - c o s t}{1 + c o s t}} = \sqrt{\frac{1 - \frac{1}{\sqrt{1 + x}}}{1 + \frac{1}{\sqrt{1 + x}}}} =$ $= \sqrt{\frac{\sqrt{1 + x} - 1}{\sqrt{1 + x} + 1}} = \sqrt{\frac{{(\sqrt{1 + x} - 1)}^{2}}{x}} = \frac{\sqrt{1 + x} - 1}{\sqrt{x}}$ $\Rightarrow I = \frac{- 4}{1 + \frac{\sqrt{1 + x} - 1}{\sqrt{x}}} + c = \frac{- 4 \sqrt{x}}{\sqrt{x} + \sqrt{1 + x} - 1} + c$ $I = F (2) - F (1) = \frac{4}{\sqrt{2}} - \frac{4 \sqrt{2}}{\sqrt{2} + \sqrt{3} - 1} =$ $= 2 \sqrt{2} + 4 \sqrt{6} + 10 + 2 \sqrt{3} + 2 \sqrt{2} = 2 (2 \sqrt{6} + \sqrt{3} + 2 \sqrt{2} + 5) . ◼$ $\frac{4 \sqrt{2}}{\sqrt{3} + \sqrt{2} - 1} \times \frac{\sqrt{3} + \sqrt{2} + 1}{\sqrt{3} + \sqrt{2} + 1} = \frac{\sqrt{3} + \sqrt{2} + 1}{2 (2 - \sqrt{6})} \times \frac{2 + \sqrt{6}}{2 + \sqrt{6}} \times 4 \sqrt{2} =$ $= - \sqrt{2} (2 \sqrt{3} + 2 \sqrt{2} + 2 + 3 \sqrt{2} + 2 \sqrt{3} + \sqrt{6}) =$ $= - \sqrt{2} (4 \sqrt{3} + 5 \sqrt{2} + \sqrt{6} + 2) = - 4 \sqrt{6} - 10 - 2 \sqrt{3} - 2 \sqrt{2}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18
	$∫_1 ^2 (dx/((√x) ×(√(x+1)) ((√(x+1)) +(√x) ))) ∫_1 ^2 (((√(x+1_( ) )) −(√x))/((√(x+1)) ×(√x)))dx ∫_1 ^2 (dx/((√x) ))−∫_1 ^2 (dx/(√(x+1))) =∣(((√x) )/(1/2))−(((√(x+1)) )/(1/2))∣_1 ^2 =2{((√2) −(√(2+1)))−((√1) −(√2) } 2((√2) −(√(3))) −((√1) −(√2) )} 2(−(√3) +2(√2) −1)$
	$\int_{1}^{2} \frac{d x}{\sqrt{x} \times \sqrt{x + 1} (\sqrt{x + 1} + \sqrt{x})}$ $\int_{1}^{2} \frac{\sqrt{x + 1_{}} - \sqrt{x}}{\sqrt{x + 1} \times \sqrt{x}} d x$ $\int_{1}^{2} \frac{d x}{\sqrt{x}} - \int_{1}^{2} \frac{d x}{\sqrt{x + 1}}$ $=∣ \frac{\sqrt{x}}{\frac{1}{2}} - \frac{\sqrt{x + 1}}{\frac{1}{2}} ∣_{1}^{2}$ $= 2 {(\sqrt{2} - \sqrt{2 + 1}) - (\sqrt{1} - \sqrt{2}}$ $2 (\sqrt{2} - \sqrt{3)} - (\sqrt{1} - \sqrt{2})}$ $2 (- \sqrt{3} + 2 \sqrt{2} - 1)$
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