By differentiating x (√(1 + x)) with respect to x , Evaluate, ∫


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Question Number 54966 by Tawa1 last updated on 15/Feb/19

$By differentiating x \sqrt{1 + x} with respect to x,$ $Evaluate, \int_{0}^{2} \frac{x}{\sqrt{1 + x}} dx$
Commented by maxmathsup by imad last updated on 17/Feb/19

$w e h a v e {(x \sqrt{1 + x})}^{^{'}} = \sqrt{1 + x} + \frac{x}{2 \sqrt{1 + x}} \Rightarrow x \sqrt{1 + x} = \int \sqrt{1 + x} + \int \frac{x}{2 \sqrt{1 + x}} d x \Rightarrow$ $\int \frac{x d x}{2 \sqrt{1 + x}} = x \sqrt{1 + x} - \int \sqrt{1 + x} d x + c \Rightarrow$ $\int_{0}^{2} \frac{x d x}{2 \sqrt{1 + x}} d x = {[x \sqrt{1 + x}]}_{0}^{2} - \int_{0}^{2} \sqrt{1 + x} d x = 2 \sqrt{3} - \int_{0}^{2} \sqrt{1 + x} d x b u t$ $\int_{0}^{2} \sqrt{1 + x} d x = \int_{0}^{2} {(1 + x)}^{\frac{1}{2}} d x = {[\frac{2}{3} {(1 + x)}^{\frac{3}{2}}]}_{0}^{2} = \frac{2}{3} (3^{\frac{3}{2}} - 1) \Rightarrow$ $I = 2 \sqrt{3} - \frac{2}{3} (\sqrt{27} - 1) = 2 \sqrt{3} - \frac{2}{3} (3 \sqrt{3} - 1) = 2 \sqrt{3} - 2 \sqrt{3} + \frac{2}{3} \Rightarrow I = \frac{2}{3} .$
Commented by maxmathsup by imad last updated on 17/Feb/19

$\frac{1}{2} I = \frac{2}{3} \Rightarrow I = \frac{4}{3} .$
Commented by Tawa1 last updated on 17/Feb/19

$God bless you sir .$
	Answered by mr W last updated on 15/Feb/19
	$((d(x(√(1+x))))/dx)=(x/(2(√(1+x))))+(√(1+x)) I_1 = ∫_( 0) ^( 2) (x/(√(1 + x))) dx I_2 = ∫_( 0) ^( 2) (√(1+x)) dx=(2/3)[(1+x)^(3/2) ]_0 ^2 =(2/3)(3(√3)−1)=2(√3)−(2/3) I=(1/2)I_1 +I_2 = ∫_( 0) ^( 2) ((x/(2(√(1 + x))))+(√(1+x))) dx=[x(√(1+x))]_0 ^2 =2(√3) ⇒I_1 =∫_( 0) ^( 2) (x/(√(1 + x))) dx=2(I_1 −I_2 )=2{2(√3)−(2(√3)−(2/3))}=(4/3)$
	$\frac{d (x \sqrt{1 + x})}{d x} = \frac{x}{2 \sqrt{1 + x}} + \sqrt{1 + x}$ $I_{1} = \int_{0}^{2} \frac{x}{\sqrt{1 + x}} dx$ $I_{2} = \int_{0}^{2} \sqrt{1 + x} dx = \frac{2}{3} {[{(1 + x)}^{\frac{3}{2}}]}_{0}^{2} = \frac{2}{3} (3 \sqrt{3} - 1) = 2 \sqrt{3} - \frac{2}{3}$ $I = \frac{1}{2} I_{1} + I_{2} = \int_{0}^{2} (\frac{x}{2 \sqrt{1 + x}} + \sqrt{1 + x}) dx = {[x \sqrt{1 + x}]}_{0}^{2} = 2 \sqrt{3}$ $\Rightarrow I_{1} = \int_{0}^{2} \frac{x}{\sqrt{1 + x}} dx = 2 (I_{1} - I_{2}) = 2 {2 \sqrt{3} - (2 \sqrt{3} - \frac{2}{3})} = \frac{4}{3}$
	Commented by Tawa1 last updated on 16/Feb/19

	$God bless you sir .$
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