 minikim38

2022-01-29

How to calculate the indefinite integral of $\frac{1}{{x}^{\frac{2}{3}}\left(1+{x}^{\frac{2}{3}}\right)}$?
I substituted,
$t=\frac{1}{{x}^{\frac{1}{3}}}$
$\frac{dt}{dx}=-\frac{1}{3{x}^{\frac{4}{3}}}$
$\frac{dt}{dx}=-\frac{{t}^{4}}{3}$
Rewriting the question,
$\int \frac{dx}{{x}^{\frac{2}{3}}+{x}^{\frac{4}{3}}}$
$-\frac{1}{3}\int \frac{dt}{{t}^{4}\left(\frac{1}{{t}^{2}}+\frac{1}{{t}^{4}}\right)}$
We have,
$-\frac{1}{3}\int \frac{dt}{{t}^{2}+1}$
$-\frac{1}{3}{\mathrm{tan}}^{-1}t+C$
$-\frac{1}{3}{\mathrm{tan}}^{-1}\left(\frac{1}{{x}^{\frac{1}{3}}}\right)+C$
$3{\mathrm{tan}}^{-1}{x}^{\frac{1}{3}}+C$
Where am I wrong? Alfred Mueller

Two observations resolve this. The first is that you didn't invert the derivative properly, so your 1/3 coefficient should be 3 because $dx=-3{t}^{-4}dt$. Secondly, $\frac{\mathrm{arctan}1}{y}=\frac{\pi }{2}-\mathrm{arctan}y$ implies there's more than one way to write the answer, with your method getting something valid. I think you were expected to substitute $t={x}^{\frac{1}{3}}$ instead; your choice does work, but it's conceptually more complex. utgyrnr0

$t=\sqrt{x},\phantom{\rule{2em}{0ex}}dt=\frac{1}{{x}^{\frac{2}{3}}}dx$
$\int \frac{1}{{x}^{\frac{2}{3}}+{x}^{\frac{4}{3}}}dx=3\int \frac{1}{{t}^{2}+1}dt$

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