Question

make a substitution to express the integrand as a rational function and then evaluate the integral. (remember to use absolute values where appropriate. use c for the constant of integration.) hint: substitute u = √6{x} ∫ dx / (√x - ∛x)

Explanation:

Step1: Express $x$ in terms of $u$

Given $u = \sqrt[6]{x}$, then $x=u^{6}$ and $dx = 6u^{5}du$.

Step2: Substitute into the integral

The integral $\int\frac{dx}{\sqrt{x}-\sqrt[3]{x}}$ becomes $\int\frac{6u^{5}du}{u^{3}-u^{2}}$.

Step3: Simplify the integrand

$\frac{6u^{5}}{u^{3}-u^{2}}=\frac{6u^{5}}{u^{2}(u - 1)}=\frac{6u^{3}}{u - 1}$.
We can perform polynomial - long division: $6u^{3}=(u - 1)(6u^{2}+6u + 6)+6$. So $\frac{6u^{3}}{u - 1}=6u^{2}+6u + 6+\frac{6}{u - 1}$.

Step4: Integrate term - by - term

$\int(6u^{2}+6u + 6+\frac{6}{u - 1})du=6\times\frac{u^{3}}{3}+6\times\frac{u^{2}}{2}+6u+6\ln|u - 1|+C$.
$=2u^{3}+3u^{2}+6u + 6\ln|u - 1|+C$.

Step5: Substitute back $u=\sqrt[6]{x}$

The result is $2x^{\frac{1}{2}}+3x^{\frac{1}{3}}+6x^{\frac{1}{6}}+6\ln|\sqrt[6]{x}-1|+C$.

Answer:

$2\sqrt{x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\ln|\sqrt[6]{x}-1|+C$