How to calculate the indefinite integral of 1x23(1+x23)? I substituted, t=1x13 dtdx=−13x43 dtdx=−t43 Rewriting the...

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

${\displaystyle t=\sqrt[3]{x},dt=\frac{1}{x^{\frac{2}{3}}}dx}$
${\displaystyle \int \frac{1}{x^{\frac{2}{3}}+x^{\frac{4}{3}}}dx=3\int \frac{1}{t^2+1}dt}$