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Question Number 72494 by aliesam last updated on 29/Oct/19

((∫x(√(x^2 +5)) dx−3∫(x/(√(x^2 +5)))dx)/(∫((x(x^2 +2))/(√(x^2 +5))) dx))

$$\frac{\int{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}\:{dx}−\mathrm{3}\int\frac{{x}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{dx}}{\int\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}\:{dx}} \\ $$

Commented by mathmax by abdo last updated on 29/Oct/19

let I=∫x(√(x^2 +5))dx , J =∫ ((xdx)/(√(x^2 +5))) K = ∫ ((x(x^2 +2))/(√(x^2 +5)))dx we have I−J =∫((x(x^2 +5)−3x)/(√(x^2 +5)))dx =∫((x^3 +5x−3x)/(√(x^2 +5)))dx =∫ ((x(x^2 +2))/(√(x^2 +5)))dx ⇒ ((I−3J)/K) =1

$${let}\:{I}=\int{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}{dx}\:\:,\:{J}\:=\int\:\:\:\frac{{xdx}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{5}}} \\ $$$${K}\:=\:\int\:\:\frac{{x}\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{5}}}{dx}\:\:{we}\:{have}\:{I}−{J}\:=\int\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{5}\right)−\mathrm{3}{x}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{5}}}{dx} \\ $$$$=\int\frac{{x}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{3}{x}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{dx}\:=\int\:\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{dx}\:\Rightarrow\:\frac{{I}−\mathrm{3}{J}}{{K}}\:=\mathrm{1} \\ $$

Answered by Tanmay chaudhury last updated on 29/Oct/19

I=((I_1 −3I_2 )/I_3 ) I_1 =∫x(√(x^2 +5)) dx =(1/2)∫(x^2 +5)^(1/2) ×d(x^2 +5) =(1/2)×(((x^2 +5)^(3/2) )/(3/2))+c_1 I_2 =∫(x/(√(x^2 +5)))dx =(1/2)∫(x^2 +5)^((−1)/2) ×d(x^2 +5) =(1/2)×(((x^2 +5)^(1/2) )/(1/2))+c_2 ∫((x(x^2 +2))/(√(x^2 +5)))dx k^2 =x^2 +5→2kdk=2xdx ∫((k^2 −5+2)/k)×kdk ∫k^2 −3 dk (k^3 /3)−3k+c_3 (((x^2 +5)^(3/2) )/3)−3(x^2 +5)^(1/2) +c_3 now pls put the values to get I...

$${I}=\frac{{I}_{\mathrm{1}} −\mathrm{3}{I}_{\mathrm{2}} }{{I}_{\mathrm{3}} } \\ $$$${I}_{\mathrm{1}} =\int{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}\:\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ×{d}\left({x}^{\mathrm{2}} +\mathrm{5}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\frac{{x}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} ×{d}\left({x}^{\mathrm{2}} +\mathrm{5}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\frac{\mathrm{1}}{\mathrm{2}}}+{c}_{\mathrm{2}} \\ $$$$\int\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{dx} \\ $$$${k}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{5}\rightarrow\mathrm{2}{kdk}=\mathrm{2}{xdx} \\ $$$$\int\frac{{k}^{\mathrm{2}} −\mathrm{5}+\mathrm{2}}{{k}}×{kdk} \\ $$$$\int{k}^{\mathrm{2}} −\mathrm{3}\:\:{dk} \\ $$$$\frac{{k}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{3}{k}+{c}_{\mathrm{3}} \\ $$$$\frac{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}−\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +{c}_{\mathrm{3}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{pls}}\:\boldsymbol{{put}}\:\boldsymbol{{the}}\:\boldsymbol{{values}}\:\boldsymbol{{to}}\:\boldsymbol{{get}}\:\boldsymbol{{I}}... \\ $$

Commented by aliesam last updated on 29/Oct/19

thank you so much sir. god bless you

$${thank}\:{you}\:{so}\:{much}\:{sir}.\:{god}\:{bless}\:{you} \\ $$$$ \\ $$

Commented by Tanmay chaudhury last updated on 29/Oct/19

most welcome sir

$${most}\:{welcome}\:{sir} \\ $$

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