Question Number 120582 by MagdiRagheb last updated on 01/Nov/20 | ||
$${Let}\:{u}={x}^{\frac{\mathrm{3}}{\mathrm{5}}} \:\:\:\:\:{du}\:=\:\frac{\mathrm{3}}{\mathrm{5}{x}^{\frac{\mathrm{2}}{\mathrm{5}}} } \\ $$$${I}\:=\:\frac{\mathrm{5}}{\mathrm{3}}\int\frac{{u}}{\:\sqrt{\mathrm{3}−\mathrm{2}{u}}}\:{du}\:=\:−\frac{\mathrm{5}}{\mathrm{6}}\:\int\frac{\mathrm{3}−\mathrm{2}{u}−\mathrm{3}}{\:\sqrt{\mathrm{3}−\mathrm{2}{u}}}\:{du} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{3}}\:\int\left[\sqrt{\mathrm{3}−\mathrm{2}{u}}\:−\mathrm{3}\left(\mathrm{3}−\mathrm{2}{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right]\:{du} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{3}}\left[−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{3}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right]+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{18}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left[−\mathrm{3}+\mathrm{2}{u}\:+\:\mathrm{9}\right]+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{18}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{6}+\mathrm{2}{u}\right)+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{9}}\sqrt{\mathrm{3}−\mathrm{2}{x}^{\frac{\mathrm{3}}{\mathrm{5}}} }\left(\mathrm{3}+{x}^{\frac{\mathrm{3}}{\mathrm{5}}} \right)+{c} \\ $$ | ||