integral of $ \int (x^3+1)^{1/3} / x^2 dx $

Question

What do you think about $$ \int \frac {(x^3+1)^{1/3}}{x^2}\, dx $$

how do you compute this ? Is it possible to use the Euler substitution?

In fact, I don't know if the integral on my sheet is $$ \int \frac {(x^3+1)^{1/3}} {x^2} \, dx $$ or $$ \int \frac{x^2 + 1}{x \sqrt {x^4 - x^2 + 1} } \frac {(x^3+1)^{1/3}} {x^2} \,dx $$

would the second expression make more sense? I thought that the second one is a little bit over complicated.

Answer 1

Hint : Divide numerator & denominator by $x$ & write the integrand as $I= \int\frac {(1+\frac1 {x^3})^{1/3}}x dx $

The whole calculus : \begin{align} I =& \int \frac{(x^3 +1)^{1/3} }{x^2} dx \\ =& \int \frac{(1+ x^{-3})^{1/3} }{x} dx \end{align}

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let $y= x^{-3} \implies dy = \frac{-1}{3} x^{-4} dx $ $$ I = \frac{-1}{3} \int \frac {(1+y)^{1/3}}ydy $$

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let $ z^3 = 1+y \implies dz =3z^2dy$

\begin{align} I &= \int \frac {z^3}{1-z^3}dz \\ &= \int \frac{z^3 + 1 - 1}{1-z^3}dz \\ &= \int \frac{z^3 - 1}{1-z^3}dz + \int \frac{ 1 }{1-z^3}dz \\ &= - \int dz + \int \frac{ 1 }{1-z^3}dz \\ however \\ \int \frac{ 1 }{z^3-1}dz &= \int \frac{1}{ (z-1)(z^2+z+1) } dz \\ &= \frac{1}{3} \int \frac{1}{z-1} dz - \frac{1}{3} \int \frac{z+2}{z^2 + z + 1} dz \\ &=\frac{1}{3} \int \frac{1}{z-1} dz - \frac{1}{6} \int \frac{2z+1}{z^2 + z + 1} dz - \frac{1}{2} \int \frac{1}{z^2 + z + 1} dz \\ &=\frac{1}{3} \int \frac{1}{z-1} dz - \frac{1}{6} \int \frac{2z+1}{z^2 + z + 1} dz - \frac{1}{2} \int \frac{1}{ (z+ \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2 } dz \\ &=\frac{1}{3} \ln( z-1 ) - \frac{1}{6} \ln(z^2 + z + 1) - \frac{1}{\sqrt 3} \arctan ( \frac{2z + 1}{ \sqrt 3} ) + cst \\ so \\ I &= -z + \frac{1}{3} \ln( z-1 ) - \frac{1}{6} \ln(z^2 + z + 1) - \frac{1}{\sqrt 3} \arctan ( \frac{2z + 1}{ \sqrt 3} ) + cst \\ i.e. \\ I &= -\sqrt[3]{1 + x^{-3} } + \frac{1}{3} \ln( \sqrt[3]{1 + x^{-3} } -1 ) - \frac{1}{6} \ln(\sqrt[3]{1 + x^{-3} } ^2 + \sqrt[3]{1 + x^{-3} } + 1) \\ &- \frac{1}{\sqrt 3} \arctan ( \frac{2\sqrt[3]{1 + x^{-3} } + 1}{ \sqrt 3} ) + cst \\ \end{align}

$$ \int\frac {(1+\frac1 {x^3})^{1/3}}x dx = -\sqrt[3]{1 + x^{-3} } + \frac{1}{3} \ln( \sqrt[3]{1 + x^{-3} } -1 ) \\ - \frac{1}{6} \ln(\sqrt[3]{1 + x^{-3} } ^2 + \sqrt[3]{1 + x^{-3} } + 1) - \frac{1}{\sqrt 3} \arctan ( \frac{2\sqrt[3]{1 + x^{-3} } + 1}{ \sqrt 3} ) + cst $$

Answer 2

Hint : Divide numerator & denominator by $x$ & write the integrand as $I= \int\frac {(1+\frac1 {x^3})^{1/3}}x dx $

Now, write $y= (1/x)^3 \implies I = (-1/3)\int \frac {(1+y)^{1/3}}ydy $

Now, write $1+y =z^3 \implies dy=3z^2dz \implies I = \int \frac {z^3}{1-z^3}dz$

From here you can use partial fraction to calculate the final integral.