Evaluate the following integrals :
`(i) int(2(x+16)^((1)/(2))+(x+16)^((3)/(4)))/(x(x+16)^((5)/(4)))dx`
`(ii) int(1+x^((1)/(2))-x^((2)/(3)))/(1+x^((1)/(3)))dx`

Explanation

`(i)` Let `(x+16)=t^(4)` so that `dx=4t^(3)dt`
Thus,
`int(2(x+16)^((1)/(2))+(x+16)^((3)/(4)))/(x(x+16)^((5)/(4)))dx=int(2t^(2)+t^(3))/((t^(4)-16)(t^(5)))*4t^(3)dt`
`=4int((t+2))/((t+2)(t-2)(t^(2)+4))dt=4int(1)/((t-2)(t^(2)+4))dt`
Now `(1)/((t-2)(t^(2)+4))=(A)/(t-2)+(Bt+C)/(t^(2)+4)`
`implies1=A(t^(2)+4)+(Bt+C)(t-2)`
Equating the coefficients of like powers of `t`, we shall have
`A=(1)/(8)`, `B=-(1)/(8)` and `C=-(1)/(4)`
Hence, `int(2(x+16)^((1)/(2))+(x+16)^((3)/(4)))/(x(x+16)^((5)/(4)))dx=(4)/(8)int((1)/(t-2)-(t+2)/(t^(2)+4))dt`
`=(1)/(2)[int(1)/(t-2)dt-(1)/(2)int(2tdt)/(t^(2)+4)-2int(1)/(t^(2)+4)dt]`
`=(1)/(2)[ln|t-2|-(1)/(2)ln(t^(2)+4)-tan^(-1)((t)/(2))+C]`
`=(1)/(2)[ln|(x+16)^((1)/(4))-2|-(1)/(2)ln((x+16)^((1)/(2))+4)-tan^(-1)(((x+16))/2)^((1)/(4))]+C`, where `C` is a constant of integration.
`(ii)` Let us put `x=t^(6)` so that `dx=6t^(5)dt` and `int(1+x^((1)/(2)-x^((2)/(3))))/(1+x^((1)/(3)))dx=int(1+t^(3)-t^(4))/(1+t^(2))xx6t^(5)dt`
`=int(1-t^(4))/(1+t^(2))*6t^(5)dt+6int(t^(8))/(1+t^(2))dt`
`=6int(t^(5)-t^(7))dt+6[int(t^(6)-t^(4)+t^(2)-1)+(1)/(1+t^(2))dt]`
`=6[(t^(6))/(6)-(t^(8))/(8)]+6[(t^(7))/(7)-(t^(5))/(5)+(t^(3))/(3)-t+tan^(-1)t]+C`
`=x-(3x^(4//3))/(4)+(6)/(7)x^(7//8)-(6)/(5)x^(5//6)+2x^(1//2)-6x^(1//6)+6tan^(-1)(x^(1//6))+C`
where `C` is a constant of integration