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I have to find the indefinite integral of $\arctan((x^2+1)^{1/3})$. I have tried to do different variable substitutions, then integration by parts, but in the end I always got a another integral that I could not calculate. Please give a hint how to calculate this integral.

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    $\begingroup$ I'm not entirely certain but there doesn't seem to be a result for the indefinite integral in terms of standard mathematical functions. $\endgroup$ Commented Mar 7, 2021 at 19:08
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    $\begingroup$ What do you mean with "have to find"? Many integrals cannot be evaluated in terms of elementary functions. This seems to be one of them. $\endgroup$
    – user
    Commented Mar 7, 2021 at 19:30
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    $\begingroup$ A CAS says it's a mess of Appell and Elliptic functions. Not likely there is a nice representation in elementary functions $\endgroup$
    – Sal
    Commented Mar 7, 2021 at 19:45

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If you do $x=y^3-1$ and $\mathrm dx=3y^2\,\mathrm dy$, then that indefinite integral becomes$$\int3y^2\arctan(y)\,\mathrm dy.$$You can deal with it by integration by parts$$\int3y^2\arctan(y)\,\mathrm dy=y^3\arctan(y)-\int\frac{y^3}{y^2+1}\,\mathrm dy.$$Can you take it from here?

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    $\begingroup$ Should not it be $x^2=y^3-1$? $\endgroup$
    – user
    Commented Mar 7, 2021 at 18:59
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Change variables $y^3 = 1+x^2$ to get $$ \int \arctan(\sqrt[3]{x^2+1}) \, dx = \frac{3}{2}\int\frac{y^2\arctan y}{\sqrt{y^3-1}}\;dy $$ Integrate by parts to get $$ {\frac {\arctan \left( y \right) \left( y-1 \right) \left( {y}^{2}+y +1 \right) }{\sqrt {{y}^{3}-1}}}-\int \!{\frac { \left( y-1 \right) \left( {y}^{2}+y+1 \right) }{ \left( {y}^{2}+1 \right) \sqrt {{y}^{3} -1}}}\,{\rm d}y \\= { {\arctan \left( y \right) \sqrt{ y^3-1 } }}-\int \!{\frac { \sqrt{y^3-1} }{ {y}^{2}+1 }}\,{\rm d}y $$ That is an elliptic integral. In Maple it is about a page long, in terms of $F$ and $\Pi$.

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Wolfram Alpha says that the result 'cannot be expressed in terms of standard mathematical functions'.

The best you can do is say that $$ \int \arctan(\sqrt[3]{x^2+1}) \, dx = \int_{0}^{x} \arctan(\sqrt[3]{t^2+1}) \, dt+C \, . $$ This result follows directly from the fundamental theorem of calculus.

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