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.) $int\frac{1}{sqrt3{x}-sqrt{x}}dx$ hint: substitute $u = sqrt6{x}$

Explanation:

Step1: Apply the substitution

Let $u = \sqrt{x}$, then $x = u^{2}$ and $dx=2u\ du$. The integral $\int\frac{dx}{\sqrt{x}-\sqrt[3]{x}}$ becomes $\int\frac{2u\ du}{u - u^{\frac{2}{3}}}$.

Step2: Simplify the integrand

Factor out $u^{\frac{2}{3}}$ from the denominator: $\int\frac{2u\ du}{u^{\frac{2}{3}}(u^{\frac{1}{3}} - 1)}=\int\frac{2u^{1-\frac{2}{3}}du}{u^{\frac{1}{3}} - 1}=\int\frac{2u^{\frac{1}{3}}du}{u^{\frac{1}{3}} - 1}$. Let $t=u^{\frac{1}{3}}$, then $u = t^{3}$ and $du = 3t^{2}dt$. The integral is now $\int\frac{2t\cdot3t^{2}dt}{t - 1}=\int\frac{6t^{3}dt}{t - 1}$.

Step3: Perform polynomial long - division

Divide $6t^{3}$ by $t - 1$. We have $6t^{3}=6t^{2}(t - 1)+6t^{2}$, then $6t^{2}=6t(t - 1)+6t$, and $6t=6(t - 1)+6$. So, $\frac{6t^{3}}{t - 1}=6t^{2}+6t + 6+\frac{6}{t - 1}$.

Step4: Integrate term - by - term

$\int(6t^{2}+6t + 6+\frac{6}{t - 1})dt=6\int t^{2}dt+6\int tdt+6\int dt+6\int\frac{dt}{t - 1}$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$) and $\int\frac{1}{x}dx=\ln|x|+C$, we get $6\cdot\frac{t^{3}}{3}+6\cdot\frac{t^{2}}{2}+6t+6\ln|t - 1|+C$.

Step5: Back - substitute

Since $t = u^{\frac{1}{3}}$ and $u=\sqrt{x}$, we have $t=x^{\frac{1}{6}}$. The result is $2x^{\frac{1}{2}}+3x^{\frac{1}{3}}+6x^{\frac{1}{6}}+6\ln|x^{\frac{1}{6}} - 1|+C$.

Answer:

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