if f'(x)show that : (x+1)ln(x+1)-1$<$$x^2$/2
okay so i want to show that 
f(x) $<$ g(x) when x$>$0
f(x)=(x+1)ln(x+1)-1
and g(x)= $x^2$/2
(x+1)ln(x+1)-1<$x^2$/2
deriving the functions give f'(x)=ln(x+1)+1 and g'(x)=x
now we want to show that f'(x)$<$g'(x) so we can integrate from 0 to x
$\displaystyle \int_{0}^{x} f'(x) \, \mathrm{d}x$<$\displaystyle \int_{0}^{x} g'(x) \, \mathrm{d}x$
and get back the original expression 
but . ln(x+1)-(1-x)=f'(x)-g'(x) isn't always negative , it's actually positive in the interval [0;2.2]
if f'(x)>g'(x) on an interval then how can the original expression be true?
we can just integrate here from 0 to 1 to obtain f(x) > g(x)
i got stuck for about an hour, this is not a homework question , don't know what i'm doing wrong here.. 
 A: Hint: It is easier if you isolate the logarithm, so it disappears when you differentiate. So rearrange the inequality to be proved into $$\ln(x+1)<\frac{1+x^2/2}{x+1},$$
let $f(x)$ be the difference of the two sides, and differentiate to find the extrema of $f$.
Addendum: As pointed out in a comment, the inequality to be proved isn't even right. But on general principles, the above advice isn't too dumb, even though it does not solve this problem. So I am letting it stand.
A: The integral $\int_0^xf'(t)dt$ on the left gives $(x + 1)\ln(x + 1)$, it does not give $(x + 1)\ln(x + 1) - 1 = f(x)$.
A: Take $h(x)=g(x)-f(x)=\frac{x^2}{2}-(x+1)\log(x+1)-1$. Note $h(0)=1>0$. This is a convex function for $x>0$ since $h''(x)=\frac{x}{x+1}>0 \ \forall x >0$. All we need to establish now is where $h(x)$ reaches its minimum for $x>0$. This is achieved by setting $h'(x)=0$ and solving for $x$, you $x^{\ast} \approx 2.14$. Note $h( x^{\ast})<0$. This means that $\ \exists \ \alpha_1< \ x^{\ast} <\alpha_2$ such that $h(x)<0 \ \forall \ \alpha_1 <x<\alpha_2$ and $h(x)>0$ for all other positive $x$. In other words, $g(x) <f(x)$ only in this interval. $\alpha_1$ and $\alpha_2$ can be found numerically. 
