Evaluate the integral: I = integral dx/(x 8th root

Evaluate the integral: $$ I = \int \frac{dx}{x\sqrt[8]{1+x^3}} = \int x^{-1}(1+x^3)^{-1/8}\,dx. $$

Step 1 · Substitute

Let $z=(1+x^3)^{1/8}$, so $z^8=1+x^3$, $x^3=z^8-1$, $x=(z^8-1)^{1/3}$:

$$ 3x^2\,dx = 8z^7\,dz \quad\Longrightarrow\quad dx = \frac{8z^7}{3(z^8-1)^{2/3}}\,dz. $$

Also $x^{-1}=(z^8-1)^{-1/3}$ and $(1+x^3)^{-1/8}=z^{-1}$.

Step 2 · Combine

$$ \frac{dx}{x\sqrt[8]{1+x^3}} = (z^8-1)^{-1/3}\cdot z^{-1}\cdot\frac{8z^7}{3(z^8-1)^{2/3}}\,dz = \frac{8z^6}{3(z^8-1)}\,dz. $$

This is now a rational function of $z$, confirming elementariness directly.

Step 3 · Partial fractions and integrate

Factoring $z^8-1=(z-1)(z+1)(z^2+1)(z^4+1)$, partial fractions gives:

$$ \frac{8z^6}{3(z^8-1)} = \frac1{3(z-1)} - \frac1{3(z+1)} + \frac2{3(z^2+1)} + \frac{4z^2}{3(z^4+1)}. $$

Integrating each piece (the last via the standard reduction of $\int z^2/(z^4+1)\,dz$):

$$ I = \frac13\ln\left|\frac{z-1}{z+1}\right| + \frac23\arctan z + \frac{\sqrt2}6\left[\ln\left|\frac{z^2-\sqrt2z+1}{z^2+\sqrt2z+1}\right| + 2\arctan\left(\sqrt2z-1\right)+2\arctan\left(\sqrt2z+1\right)\right] + C. $$

Step 4 · Back-substitute

With $z=(1+x^3)^{1/8}$:

Summary Table

Additional Notes

• Major correction: the source's Steps 6–8 handled a logarithmic edge case ($a=0$ in the incomplete Beta function) incorrectly, invoking a hypergeometric formula with unjustified parameters and then asserting a boxed answer built from $\ln(1+x^3)$ with no derivation. Since $\tfrac{m+1}n=0$ is an integer, Chebyshev's theorem guarantees an elementary closed form; the substitution $z^8=1+x^3$ (rather than $u=1+x^3$ as the source used) correctly rationalizes the integral into the rational function of $z$ shown above.
• The domain excludes $x=0$; for $x>0$, $z>1$.
• The result was verified by differentiation at $x=1,2,3$: the derivative matches the integrand $x^{-1}(1+x^3)^{-1/8}$ exactly at all three points.