$\int \frac{1}{x^{1/2}+x^{1/3}}dx=A\sqrt{x}-B{x}^{1/3}+C{x}^{1/6}+D\log |x^{1/6}+1|+K$ where ‘K’ is constant of integration then the value of A+B+C+D is

$$\begin{gathered} I=\int \frac{1}{{\mathrm{x}}^{1/2}+{\mathrm{x}}^{1/3}}\mathrm{dx} \\ \mathrm{Put} \mathrm{x}={\mathrm{t}}^6 \\ \mathrm{dx}=6{\mathrm{t}}^5 \mathrm{dt} \\ \mathrm{I}=\int \frac{1}{{\mathrm{t}}^3+{\mathrm{t}}^2}\times 6 {\mathrm{t}}^5 \mathrm{dt} \\ =6\int \frac{{\mathrm{t}}^3}{\mathrm{t}+1}\mathrm{dt} \\ =6\int \frac{({\mathrm{t}}^3+1)-1}{\mathrm{t}+1}\mathrm{dt} \\ =6\int \left(\left({\mathrm{t}}^2+\mathrm{t}+1\right)-\frac{1}{\mathrm{t}+1}\right)\mathrm{dt} \\ =6\left[\frac{{\mathrm{t}}^3}{3}+\frac{{\mathrm{t}}^2}{2}+\mathrm{t}-\log (\mathrm{t}+1)\right] \\ =2 {\mathrm{x}}^{1/2}+3 {\mathrm{x}}^{1/3}+6{\mathrm{x}}^{1/6}-6 \log ({\mathrm{x}}^{1/6}+1) \\ \mathrm{A}=2, \mathrm{B}=-3, \mathrm{C}=6, \mathrm{D}=-6 \\ \mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D}=-1 \end{gathered}$$