Taylor's Remainder $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) Prove that 
$\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) <x-\frac{x^2}{2}+\frac{x^3}{3}$
My attempt: I can prove this by taking one side at a time and assuming $\displaystyle f(x) = \log (1+x)-x+\frac{x^2}{2}-\frac{x^3}{3}$ 
and then proving that it is a decreasing function and then the same for the other side.
But I am looking for a better solution using Lagrange's Mean Value Theorem or using Taylor's remainder.

This is shown in
section 14 of
Heinrich Dorrie's
"100 Great Problems of
Elementary Mathematics",
which I highly recommend.
The proof,
for general $n$,
goes like this:
Let
$f(t) = \dfrac1{1+t}$.
Then,
as can be shown by induction,
$$f(t)
=\sum_{k=0}^{n-1} (-1)^{k}t^k+(-1)^nt^nf(t)
$$
Integrating from
$0$ to $x$,
\begin{align}\\
\ln(1+x)
&=\int_0^x \frac{dt}{1+t}\\
&=\int_0^x \left(\sum_{k=0}^{n-1} (-1)^{k}t^k+(-1)^nt^nf(t)\right)dt\\
&=\sum_{k=0}^{n-1} \int_0^x(-1)^{k}t^kdt+\int_0^x(-1)^nt^nf(t)dt\\
&=\sum_{k=0}^{n-1} (-1)^{k}\frac{x^{k+1}}{k+1}+(-1)^n\int_0^x\frac{t^ndt}{1+t}\\
&=\sum_{k=0}^{n-1} (-1)^{k}\frac{x^{k+1}}{k+1}+(-1)^nr_n(t)\\
\end{align}
where
$$r_n(t)
=\int_0^x\frac{t^ndt}{1+t}
$$
Since
$1 \le 1+t
\le 1+x
$ for
$0 \le t \le x$,
$$t^n 
\ge \frac{t^n}{1+t}
\ge \frac{t^n}{1+x}
$$
so that,
since
$$\int_0^x t^n dt
=\frac{x^{n+1}}{n+1}
$$
$$\frac{x^{n+1}}{n+1}
\ge r_n(x)
\ge \frac{x^{n+1}}{(n+1)(1+x)}
$$
Therefore
$$r_n(x)
=\frac{x^{n+1}}{(n+1)(1+v)}
$$
where
$0 < v < x
$.
Therefore
$$\ln(1+x)
=\sum_{k=0}^{n-1} (-1)^{k}\frac{x^{k+1}}{k+1}
+(-1)^n\frac{x^{n+1}}{(n+1)(1+v)}
$$
where
$0 <v < x$.
