Question Number 22545 by vajpaithegrate@gmail.com last updated on 20/Oct/17
$$\int\frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }\mathrm{dx}= \\ $$$$ \\ $$
Answered by $@ty@m last updated on 20/Oct/17
$${Its}\:{similar}\:{to}\:{Q}.\:{No}.\:\mathrm{20540} \\ $$
Commented by vajpaithegrate@gmail.com last updated on 20/Oct/17
$$\mathrm{Thank}\:\mathrm{u}\:\mathrm{sir} \\ $$
Answered by sma3l2996 last updated on 20/Oct/17
$${I}=\int\frac{\sqrt{{x}}}{\sqrt{{x}}−\sqrt[{\mathrm{3}}]{{x}}}{dx}=\int\frac{{x}}{{x}−\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}{dx} \\ $$$${u}=\sqrt[{\mathrm{3}}]{{x}}\Leftrightarrow{x}={u}^{\mathrm{3}} \Rightarrow{dx}=\mathrm{3}{u}^{\mathrm{2}} {du} \\ $$$${I}=\int\frac{{u}^{\mathrm{3}} }{{u}^{\mathrm{3}} −{u}^{\mathrm{2}} }\left(\mathrm{3}{u}^{\mathrm{2}} {du}\right) \\ $$$${I}=\mathrm{3}\int\frac{{u}^{\mathrm{3}} }{{u}−\mathrm{1}}{du}=\mathrm{3}\int\frac{{u}^{\mathrm{3}} −\mathrm{1}+\mathrm{1}}{{u}−\mathrm{1}}{du} \\ $$$$=\mathrm{3}\int\frac{\left({u}−\mathrm{1}\right)\left({u}^{\mathrm{2}} +{u}+\mathrm{1}\right)}{{u}−\mathrm{1}}{du}+\mathrm{3}\int\frac{{du}}{{u}−\mathrm{1}} \\ $$$$=\mathrm{3}\int\left({u}^{\mathrm{2}} +{u}+\mathrm{1}\right){du}+\mathrm{3}{ln}\mid{u}−\mathrm{1}\mid+{c} \\ $$$${I}={u}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} +\mathrm{3}{u}+\mathrm{3}{ln}\mid{u}−\mathrm{1}\mid+{C} \\ $$$${I}={x}+\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }+\mathrm{3}\sqrt[{\mathrm{3}}]{{x}}+\mathrm{3}{ln}\mid\sqrt[{\mathrm{3}}]{{x}}−\mathrm{1}\mid+{C} \\ $$
Commented by vajpaithegrate@gmail.com last updated on 21/Oct/17
$$\int\frac{\sqrt{\mathrm{x}}}{\sqrt{}\mathrm{x}−^{\mathrm{3}} \sqrt{\mathrm{x}}}\mathrm{dx}=\int\frac{\mathrm{x}}{\mathrm{x}−^{\mathrm{3}} \sqrt{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}\:\mathrm{is}\:\mathrm{it}\:\mathrm{correc} \\ $$
Commented by $@ty@m last updated on 21/Oct/17
$$\:^{\mathrm{3}} \sqrt{{x}}×\sqrt{{x}} \\ $$$$={x}^{\frac{\mathrm{1}}{\mathrm{3}}} ×{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$={x}^{\frac{\mathrm{5}}{\mathrm{6}}} \\ $$$$\:\:^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }={x}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\therefore\:\:^{\mathrm{3}} \sqrt{{x}}×\sqrt{{x}}\neq\:^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} } \\ $$
Commented by vajpaithegrate@gmail.com last updated on 21/Oct/17
$$\mathrm{yes}\:\mathrm{sir} \\ $$