Question Number 120582 by MagdiRagheb last updated on 01/Nov/20
![Let u=x^(3/5) du = (3/(5x^(2/5) )) I = (5/3)∫(u/( (√(3−2u)))) du = −(5/6) ∫((3−2u−3)/( (√(3−2u)))) du = −(5/3) ∫[(√(3−2u)) −3(3−2u)^(−(1/2)) ] du = −(5/3)[−(1/3)(3−2u)^(3/2) +3(3−2u)^(1/2) ]+c = −(5/(18))(3−2u)^(1/2) [−3+2u + 9]+c = −(5/(18))(3−2u)^(1/2) (6+2u)+c = −(5/9)(√(3−2x^(3/5) ))(3+x^(3/5) )+c](https://www.tinkutara.com/question/Q120582.png)
$${Let}\:{u}={x}^{\frac{\mathrm{3}}{\mathrm{5}}} \:\:\:\:\:{du}\:=\:\frac{\mathrm{3}}{\mathrm{5}{x}^{\frac{\mathrm{2}}{\mathrm{5}}} } \\ $$$${I}\:=\:\frac{\mathrm{5}}{\mathrm{3}}\int\frac{{u}}{\:\sqrt{\mathrm{3}−\mathrm{2}{u}}}\:{du}\:=\:−\frac{\mathrm{5}}{\mathrm{6}}\:\int\frac{\mathrm{3}−\mathrm{2}{u}−\mathrm{3}}{\:\sqrt{\mathrm{3}−\mathrm{2}{u}}}\:{du} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{3}}\:\int\left[\sqrt{\mathrm{3}−\mathrm{2}{u}}\:−\mathrm{3}\left(\mathrm{3}−\mathrm{2}{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right]\:{du} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{3}}\left[−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{3}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right]+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{18}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left[−\mathrm{3}+\mathrm{2}{u}\:+\:\mathrm{9}\right]+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{18}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{6}+\mathrm{2}{u}\right)+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{9}}\sqrt{\mathrm{3}−\mathrm{2}{x}^{\frac{\mathrm{3}}{\mathrm{5}}} }\left(\mathrm{3}+{x}^{\frac{\mathrm{3}}{\mathrm{5}}} \right)+{c} \\ $$