Let $\text{I}=\int\frac{1}{\text{x}^{\frac{1}{3}}\big(\text{x}^{\frac{1}{3}}-1\big)}\text{dx}$
$=\int\frac{1}{\text{x}^{\frac{2}{3}}-\text{x}^{\frac{1}{3}}}\text{dx}$
Let $\text{x}=\text{t}^{3}$
On differentiating both sides, we get
$\text{dx}=3\text{t}^{2}\text{dt}$
$\therefore\ \text{I}\int\frac{3\text{t}^{2}}{(\text{t})^{\frac{2}{3}}-(\text{t}^{3})^{\frac{1}{3}}}\text{dt}$
$=\int\frac{3\text{t}^{2}}{\text{t}^2-\text{t}}\text{dt}$
$=3\int\frac{\text{t}}{\text{t}-1}\text{dt}$
$=3\int\frac{(\text{t}-1)+1}{\text{t}-1}\text{dt}$
$=3\int\Big[(1)+\frac{1}{\text{t}-1}\Big]\text{dt}$
$=\big[1+\log(\text{t}-1)\big]+\text{C}$
$=3\text{x}^\frac{1}{3}+3\log\big({\text{x}^\frac{1}{3}-1\big)}+\text{C}$
Hence, $\int\frac{1}{\text{x}^{\frac{1}{3}}\big(\text{x}^{\frac{1}{3}}-1\big)}\text{dx}=3\text{x}^\frac{1}{3}+3\log\big({\text{x}^\frac{1}{3}-1\big)}+\text{C}$
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