Question Number 164376 by muneer0o0 last updated on 16/Jan/22
Answered by alephzero last updated on 16/Jan/22
$$\int\frac{\mathrm{d}{x}}{\:\sqrt{{x}^{\mathrm{3}} +\mathrm{64}}}\:=\:? \\ $$$${x}^{\mathrm{3}} \:=\:\left(\sqrt{{x}^{\mathrm{3}} }\right)^{\mathrm{2}} \\ $$$$\int\frac{\mathrm{d}{u}}{\:\sqrt{{u}^{\mathrm{2}} +{a}}}\:=\:\mathrm{ln}\:\mid{u}+\sqrt{{u}^{\mathrm{2}} +{a}}\mid+\mathrm{C} \\ $$$$\Rightarrow\:\int\frac{\mathrm{d}{x}}{\:\sqrt{\left(\sqrt{{x}^{\mathrm{3}} }\right)^{\mathrm{2}} +\mathrm{64}}}\:=\:\mathrm{ln}\:\mid\sqrt{{x}^{\mathrm{3}} }+\sqrt{{x}^{\mathrm{3}} +\mathrm{64}}\mid+\mathrm{C} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{wrong}. \\ $$$$\left(\mathrm{I}\:\mathrm{just}\:\mathrm{started}\:\mathrm{learn}\:\mathrm{calculus}\right). \\ $$
Commented by mr W last updated on 16/Jan/22
$${yes},\:{very}\:{wrong}!\:{it}'{s}\:{even}\:{not}\:{integrable} \\ $$$${using}\:{elementary}\:{functions}. \\ $$
Answered by ajfour last updated on 16/Jan/22
$${I}=\int\frac{{dx}}{\left({x}^{\mathrm{3}} +{a}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{2}} } \\ $$$${let}\:\:\:{x}={a}\left(\mathrm{tan}\:^{\mathrm{2}} \theta\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${dx}=\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta\right){d}\theta}{\mathrm{3}\left(\mathrm{tan}\:\theta\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$${I}=\int\frac{\mathrm{2}{d}\theta}{\mathrm{3}{a}\sqrt{{a}}\left(\mathrm{tan}\:\theta\right)^{\mathrm{1}/\mathrm{3}} \mathrm{cos}\:\theta} \\ $$$$\:\:\:=\frac{\mathrm{2}}{\mathrm{3}{a}\sqrt{{a}}}\int\frac{{d}\theta}{\mathrm{sin}\:^{\mathrm{1}/\mathrm{3}} \theta\mathrm{cos}\:^{\mathrm{2}/\mathrm{3}} \theta} \\ $$$${I}\:{think}\:{now}\:{Gamma}\:{function} \\ $$$${should}\:{be}\:{used}... \\ $$