`int1/(x^(1/2)+x^(1/3))dx`

Explanation

`" [Hint]" :(1)/(x^(1/2)+x^(1/3))=(1)/(x^(1/3)(1+x^(1/6))), " put " x=t^(6)"]"`
`" Let " I= int (1)/(x^(1//2)+x^(1//3))dx`
`" Let " x=t^(6) rArr dx=6t^(6) dt`
` :. I= int(6t^(5))/((t^(6))^(1//2)+(t^(6))^(1//3))dt=6 int (t^(5))/(t^(3)+t^(2))dt`
`=6int (t^(5))/(t^(2)(t+1))dt=6 int (t^(3))/((t+1))dt`
`=6 int ((t^(3)+1-1)/(t+1)) dt`
`=6 int [((t+1)(t^(2)-t+1))/(t+1)-(1)/(t+1)]dt`
`=6int (t^(2)-5+1-(1)/(t+1))dt`
`=6((t^(3))/(3)-(t^(2))/(2)+t-log|t+1|)+C`
`=6 {((x^(1//6))^(3))/(3)-((x^(1//6))^(2))/(2) +(x^(1//6))-log |x^(1//6)+1|+C}`
`=2x^(1//2) -3x^(1//3)+6x^(1//6) -6log |x^(1//6)+1|+C`