Question Number 201110 by emilagazade last updated on 29/Nov/23
$$\int\frac{\mathrm{1}}{\:\sqrt{\left({x}−{a}\right)^{\mathrm{3}} }+\sqrt{\left({x}+{a}\right)^{\mathrm{3}} }}{dx} \\ $$
Answered by Frix last updated on 29/Nov/23
$$\sqrt{{p}}+\sqrt{{q}}=\sqrt{{p}+\mathrm{2}\sqrt{{pq}}+{q}} \\ $$$$\int\frac{{dx}}{\:\left({x}−{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dx}}{\left({x}\left({x}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \right)+\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} }= \\ $$$$\left(\mathrm{use}\:{t}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}\:\Leftrightarrow\:{x}=\frac{{a}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}{t}}\:\rightarrow\:{dx}=\frac{{a}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{t}^{\mathrm{2}} }{dt}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{a}}}\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({t}^{\mathrm{2}} +\mathrm{3}\right)}{dt}\:\overset{{u}=\sqrt{{t}}} {=}\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{{a}}}\int\frac{{u}^{\mathrm{4}} −\mathrm{1}}{{u}^{\mathrm{2}} \left({u}^{\mathrm{4}} +\mathrm{3}\right)}{du}= \\ $$$$\left(\mathrm{decompose}…\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{{a}}{u}}+\frac{\mathrm{2}}{\:\mathrm{3}^{\frac{\mathrm{5}}{\mathrm{4}}} \sqrt{{a}}}\left(\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\sqrt{\mathrm{2}}{u}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{4}}} }+\mathrm{1}\right)\:+\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\sqrt{\mathrm{2}}{u}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{4}}} }−\mathrm{1}\right)\right)+\frac{\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{5}}{\mathrm{4}}} \sqrt{{a}}}\mathrm{ln}\:\frac{{u}^{\mathrm{2}} −\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{4}}} \sqrt{\mathrm{2}}{u}+\sqrt{\mathrm{3}}}{{u}^{\mathrm{2}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{4}}} \sqrt{\mathrm{2}}{u}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Now}\:\mathrm{insert} \\ $$$${u}=\frac{\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}}{\:\sqrt{{a}}} \\ $$
Commented by emilagazade last updated on 29/Nov/23
$${thank}\:{you}\:{a}\:{lot}\:{Sir} \\ $$
Commented by Frix last updated on 29/Nov/23
$$\mathrm{I}\:\mathrm{also}\:\mathrm{tried} \\ $$$$\frac{\mathrm{1}}{\left({x}−{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\frac{\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left({x}−{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{2}{a}\left(\mathrm{3}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)} \\ $$$$\mathrm{leading}\:\mathrm{to} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{a}}\left(\int\frac{\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}−\int\frac{\left({x}−{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}\right) \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{better}… \\ $$
Commented by emilagazade last updated on 29/Nov/23
$${this}\:{one}\:{I}\:{tried}\:{too}.\:{but}\:{the}\:{book}\:{suggests}\:{to}\:{do}\:{substitution}\:{which}\:{leads}\:{to}\:{partial}\:{fractions}. \\ $$