Question

Intigretion of 1/ sqrt x + qube x

Answer 1

$\displaystyle\int\frac{\mathrm dx}{x^{1/2}+x^{1/3}}$

Let $u=x^{1/6}$, so that $u^6=x$ and $6u^5\,\mathrm du=\mathrm dx$. Then $x^{1/2}=x^{3/6}=u^3$ and $x^{1/3}=x^{2/6}=u^2$.

$\displaystyle\int\frac{6u^5}{u^3+u^2}\,\mathrm du=6\int\frac{u^3}{u+1}\,\mathrm du$

Then with $t=u+1$, so that $u=t-1$ and $\mathrm du=\mathrm dt$, we have

$\displaystyle6\int\frac{(t-1)^3}t\,\mathrm dt=6\int\left(t^2-3t+3-\frac1t\right)\,\mathrm dt$

which should be easy to finish.