gorgeousgen9487

2022-07-07

Equation:

isscacabby17

Expert

I think it's easier to make a substitution:
$u=\sqrt{{y}^{3}-2}⇒{u}^{2}={y}^{3}-2⇒2udu=3{y}^{2}dy$
$\begin{array}{rl}\int {y}^{5}\sqrt{{y}^{3}-2}dy& =\frac{1}{3}\int {y}^{3}\sqrt{{y}^{3}-2}\left(3{y}^{2}dy\right)\\ & =\frac{1}{3}\int \left({u}^{2}+2\right)u\left(2udu\right)\\ & =\frac{2}{3}\int {u}^{4}+2{u}^{2}du\\ & =\frac{2}{3}\left(\frac{{u}^{5}}{5}+2\frac{{u}^{3}}{3}\right)+C\\ & =\frac{2}{15}\left({y}^{3}-2{\right)}^{5/2}+\frac{4}{9}\left({y}^{3}-2{\right)}^{3/2}+C\end{array}$

Grimanijd

Expert

Now,

$=\frac{2}{15}{x}^{\frac{5}{2}}+\frac{4}{9}{x}^{\frac{3}{2}}=\frac{2}{15}\left({y}^{3}-3{\right)}^{\frac{5}{2}}+\frac{4}{9}\left({y}^{3}-3{\right)}^{\frac{3}{2}}.$

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