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Equation: $$\int y^5 \sqrt{y^3 - 2}\ \mathrm dy$$

I'm trying to answer this equation using the integration by parts method, how should I approach this, should I rewrite or simplify the equation first before I proceed? if yes, someone please help me, an answer on this equation and as well an explanation would be so much appreciated too. Thanks and advance and have a nice day.

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2 Answers 2

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You are thinking of doing it by integration by parts, but I think it would not be a good option, I recommend you to think about substitution before any method and in case you do not see an exit with this method, just think about integration by parts. I think it's easier to make a substitution: $$u=\sqrt{y^3-2}\Rightarrow u^2=y^3-2\Rightarrow 2udu=3y^2dy$$

\begin{align} \int y^5 \sqrt{y^3 - 2} dy&=\frac{1}{3}\int y^3\sqrt{y^3-2} (3y^2dy)\\ &=\frac{1}{3}\int(u^2+2)u(2udu)\\ &=\frac{2}{3}\int u^4+2u^2du\\ &=\frac{2}{3}\left(\frac{u^5}{5}+2\frac{u^3}{3}\right)+C\\ &=\frac{2}{15}(y^3-2)^{5/2}+\frac{4}{9}(y^3-2)^{3/2}+C \end{align}

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    $\begingroup$ Thanks mate! appreciate your clear explanation on this, Yeah, It is easier to use the u-substitution method on this one. $\endgroup$
    – Jerwell
    Commented Mar 3, 2022 at 1:22
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$$y^3-2=x \implies \int y^5 \sqrt{y^3 - 2}\ \mathrm dy =\frac13\int (x+2) \sqrt{x}\ \mathrm dx. $$ Now, $$\frac13\int (x+2) \sqrt{x}\ \mathrm dx=\frac13\int x \sqrt{x} \ \mathrm dx+\frac13\int2 \sqrt{x}\ \mathrm dx$$$$=\frac2{15}x^{\frac52}+\frac49x^{\frac32}=\frac2{15}(y^3-3)^{\frac52}+\frac49(y^3-3)^{\frac32}. $$

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    $\begingroup$ typo:$y^2$ should be $y^3$. $\endgroup$
    – Lai
    Commented Mar 17, 2022 at 10:11

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