find-dx-x-x-1-x-2-

Question Number 37630 by math khazana by abdo last updated on 16/Jun/18

$${find}\:\:\int\:\:\:\:\:\:\:\:\frac{{dx}}{\:\sqrt{{x}}\:\:+\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}+\mathrm{2}}} \\ $$

Answered by Ahmed Neutron last updated on 16/Jun/18

∫(((√x)/dx))^(−1) +∫(((√(x+1))/dx))^(−1) +∫(((√(x+2))/dx)) ∫((1/x))^(1/2) dx+∫((1/(x+1)))^(1/2) dx+∫((1/(x+2)))^(1/2) (3/2)[((1/x))^(3/2) +((1/(x+1)))^(3/2) +((1/(x+1)))^(3/2) ]+constant

$$\int\left(\frac{\sqrt{{x}}}{{dx}}\right)^{−\mathrm{1}} +\int\left(\frac{\sqrt{{x}+\mathrm{1}}}{{dx}}\right)^{−\mathrm{1}} +\int\left(\frac{\sqrt{{x}+\mathrm{2}}}{{dx}}\right)\:\: \\ $$$$\int\left(\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx}+\int\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx}+\int\left(\frac{\mathrm{1}}{{x}+\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\left[\left(\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]+{constant} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18

$${do}\:{differentiating}\:{the}\:{results}\:{give}\:{the}\:{intregation} \\ $$

Commented by MJS last updated on 16/Jun/18

it′s totally wrong (((√x)/dx))^(−1) +(((√(x+1))/dx))^(−1) +(((√(x+2))/dx))^(−1) = =(dx/( (√x)))+(dx/( (√(x+1))))+(dx/( (√(x+2))))= =(((√(x(x+1)))+(√(x(x+2)))+(√((x+1)(x+2))))/( (√(x(x+1)(x+2)))))dx≠ ≠(dx/( (√x)+(√(x+1))+(√(x+2)))) x=2: (((√6)+(√8)+(√(12)))/( (√(24))))=(1/2)+((√2)/2)+((√3)/3)≈1.7845 (1/( (√2)+(√3)+(√4)))=(2/(23))+((5(√2))/(23))+((3(√3))/(23))−((4(√6))/(23))≈.19432

$$\mathrm{it}'\mathrm{s}\:\mathrm{totally}\:\mathrm{wrong} \\ $$$$\left(\frac{\sqrt{{x}}}{{dx}}\right)^{−\mathrm{1}} +\left(\frac{\sqrt{{x}+\mathrm{1}}}{{dx}}\right)^{−\mathrm{1}} +\left(\frac{\sqrt{{x}+\mathrm{2}}}{{dx}}\right)^{−\mathrm{1}} = \\ $$$$=\frac{{dx}}{\:\sqrt{{x}}}+\frac{{dx}}{\:\sqrt{{x}+\mathrm{1}}}+\frac{{dx}}{\:\sqrt{{x}+\mathrm{2}}}= \\ $$$$=\frac{\sqrt{{x}\left({x}+\mathrm{1}\right)}+\sqrt{{x}\left({x}+\mathrm{2}\right)}+\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}}{\:\sqrt{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}}{dx}\neq \\ $$$$\neq\frac{{dx}}{\:\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}} \\ $$$${x}=\mathrm{2}:\:\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{8}}+\sqrt{\mathrm{12}}}{\:\sqrt{\mathrm{24}}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\approx\mathrm{1}.\mathrm{7845} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}}=\frac{\mathrm{2}}{\mathrm{23}}+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{23}}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{23}}−\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{23}}\approx.\mathrm{19432} \\ $$

Commented by prof Abdo imad last updated on 16/Jun/18

$${how}\:{do}\:{you}\:{get}\:{this}\:{equality}\:?\:{its}\:{not}\:{correct} \\ $$$${sir}\:{Ahmed}… \\ $$

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