Question

¿Integral de 1/[x^(1/2)+x^(1/3)]?

Answer 1

$\displaystyle I=\int \dfrac{dx}{\sqrt{x}+\sqrt[3]x}\\ \\ \text{Sea }x=u^6\to dx =6u^5du\\ \\ I=\int\dfrac{6u^5du}{|u|^3+u^2}\\ \\ I=\int\dfrac{6u^3du}{|u|+1}\\ \\ \text{Si } u\ \textless \ 0\text{ entonces}\\ \\ I_1=-6\int \dfrac{u^3du}{u-1}$

$\displaystyle I_1=-6\int u^2+u+1+\dfrac{1}{u-1}du\\ \\ I_1=-6\left[\dfrac{u^3}{3}+\dfrac{u^2}{2}+u+\ln(1-u)\right]\\ \\ \\ \boxed{I_1=-2u^3-3u^2-6u-6\ln(1-u)}$

Si $u\geq 0 \text{ entonces}$

$\displaystyle I_2=6\int \dfrac{u^3}{u+1}du\\ \\ I_2=6\int u^2-u+1-\dfrac{1}{u+1}du\\ \\ I_2=6\left[\frac{u^3}{3}-\frac{u^2}{2}+u-\ln(u+1)\right]\\ \\\\ \boxed{I_2=2u^3-3u^2+6u-6\ln(u+1)}$

Por consiguiente:
$I=2|u|^3-3u^2+6|u|-6\ln(|u|+1)+C\\ \\ I=2|\sqrt[6]x|^3-3\sqrt[6]x^2+6|\sqrt[6]x|-6\ln(|\sqrt[6]x|+1)+C\\ \\ \\ \boxed{\boxed{I=2\sqrt{x}-3\sqrt[3]x+6\sqrt[6]x-6\ln(\sqrt[6]x+1)+C}}$