Prove that the following functions are strictly increasing: `f(x)=log(1+x)-(2x)/(2+x)` for x > -1

Explanation

f(X)=`cot^(-1)x+x`
Differentiating w.r.t x we get
`f(X)=(-1)/(1+x^(2))=(-1+1+x^(2))/(1+x^(2))=(x^(2))/(1+x^(2))`
Clearly, `f(X) ge 0 forall (f(x)=0 for x =0` only)
So f(X) increases in `(-oo,oo)`
(b) `f(X) =log(1+x)-(2x)/(2+x)`
`therefore f(X)=(1)/(1+x)-(2(2+x)-2x)/(2+x)^(2)`
`=(x^(2))/(x+1)(x+2)^(2)`
Obviously ,`f(x) ge 0 forall x gt -1 f(x)=0` for x =0 only
Hence f(X) is increasing on `(-1,oo)`