Question Number 72494 by aliesam last updated on 29/Oct/19
$$\frac{\int{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}\:{dx}−\mathrm{3}\int\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{dx}}{\int\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}\:{dx}} \\ $$
Commented by mathmax by abdo last updated on 29/Oct/19
$${let}\:{I}=\int{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}{dx}\:\:,\:{J}\:=\int\:\:\:\frac{{xdx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{5}}} \\ $$$${K}\:=\:\int\:\:\frac{{x}\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{5}}}{dx}\:\:{we}\:{have}\:{I}−{J}\:=\int\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{5}\right)−\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{5}}}{dx} \\ $$$$=\int\frac{{x}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{dx}\:=\int\:\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{dx}\:\Rightarrow\:\frac{{I}−\mathrm{3}{J}}{{K}}\:=\mathrm{1} \\ $$
Answered by Tanmay chaudhury last updated on 29/Oct/19
$${I}=\frac{{I}_{\mathrm{1}} −\mathrm{3}{I}_{\mathrm{2}} }{{I}_{\mathrm{3}} } \\ $$$${I}_{\mathrm{1}} =\int{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}\:\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ×{d}\left({x}^{\mathrm{2}} +\mathrm{5}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} ×{d}\left({x}^{\mathrm{2}} +\mathrm{5}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\frac{\mathrm{1}}{\mathrm{2}}}+{c}_{\mathrm{2}} \\ $$$$\int\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{dx} \\ $$$${k}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{5}\rightarrow\mathrm{2}{kdk}=\mathrm{2}{xdx} \\ $$$$\int\frac{{k}^{\mathrm{2}} −\mathrm{5}+\mathrm{2}}{{k}}×{kdk} \\ $$$$\int{k}^{\mathrm{2}} −\mathrm{3}\:\:{dk} \\ $$$$\frac{{k}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{3}{k}+{c}_{\mathrm{3}} \\ $$$$\frac{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}−\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +{c}_{\mathrm{3}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{pls}}\:\boldsymbol{{put}}\:\boldsymbol{{the}}\:\boldsymbol{{values}}\:\boldsymbol{{to}}\:\boldsymbol{{get}}\:\boldsymbol{{I}}… \\ $$
Commented by aliesam last updated on 29/Oct/19
$${thank}\:{you}\:{so}\:{much}\:{sir}.\:{god}\:{bless}\:{you} \\ $$$$ \\ $$
Commented by Tanmay chaudhury last updated on 29/Oct/19
$${most}\:{welcome}\:{sir} \\ $$