evaluate the indefinite integral please $$\int\left(\frac{dx}{\sqrt[3]x+\sqrt{x}}\right)$$ No ideas how to evaluate it
2026-04-24 18:21:34.1777054894
Integral $\int(\frac{dx}{\sqrt[3]x+\sqrt{x}})$
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Hint: if we take the substitution $u=x^{1/6} $ we get $$\int\frac{dx}{\sqrt[3]{x}+\sqrt{x}}=6\int\frac{u^{3}}{u+1}du $$ then, use the identity $$\frac{u^{3}}{u+1}=u^{2}-u-\frac{1}{u+1}+1.$$