Question

Intigretion of 1/ sqrt x + qube x

Answer 1

`\displaystyle\int(\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(6u^5)/(u^3+u^2)\,\mathrm du=6\int(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.