Prove equality with an integral without the series expansion of $\ln(1+x)$ 
Let $P_n(x)=\displaystyle\sum_{k=1}^{2n}   \frac{(-1)^kx^k}k$ for all $x\in [0,+\infty)$
Show that, for all $n$ integer,
  $$
P_n(x)=\int_{0}^{x} \frac{t^{2n}-1}{t+1} dt.
$$

Can I solve this problem without :
$$
\ln(1+x)=\sum_{k=1}^{+\infty} \frac{(-1)^{k+1}x^k}k 
$$
Actually I am helping a friend which does not know the series expansion of $\ln(1+x)$.
Unfortunately, I have not managed to prove differently.
Thank you in advance,
EDIT: My approach doesn't work because  $x\in [0,+\infty)$
 A: We have $$\frac{t^{2n}-1}{1+t}=(t^{2n}-1)(1-t+t^2-t^3+...)=(t^{2n}-...)-(1-t+...)=1-t+...+t^{2n-1}$$ Integrating this gives $$x-{x^2\over 2}+...+{x^{2n}\over {2n}}$$ as desired.
EDIT: Sorry, I have indeed switched the signs. Thank you for alerting me to this.
A: This can be done using induction: Suppose it works for some $n$, and note that
\begin{align*}
P_{n + 1}(x) &= \sum_{k = 1}^{2n + 2} \frac{(-1)^k x^k}{k} \\
&= P_n(x) - \frac{x^{2n + 1}}{2n + 1} + \frac{x^{2n + 2}}{2n + 2} \\
&= \int_0^x \frac{t^{2n} - 1}{t + 1} dt - \int_0^x t^{2n} dt + \int_0^x t^{2n + 1} dt \\
&= \int_0^x \frac{t^{2n} - t^{2n} (t + 1) + t^{2n + 1}(t + 1)}{t + 1} dt \\
&= \int_0^x \frac{t^{2n} - t^{2n + 1} - t^{2n} + t^{2n + 2} + t^{2n + 1}}{t + 1} dt \\
&= \int_0^x \frac{t^{2(n + 1)} - 1}{t + 1} dt
\end{align*}
as desired. (One still must do the base case, of course.)
A: Notice that $\frac{t^{2n}-1}{t+1}=-\frac{(t+1)(1-t+t^2-t^3+...-t^{2n-1})}{t+1}=-1+t-t^2+t^3-...+t^{2n-1}$
Thus $\int_0^x \frac{t^{2n}-1}{t+1}dt= \int_0^x -1+t-t^2+t^3-...+t^{2n-1}dt=(-t+\frac{1}{2}t^2-...+\frac{1}{2n}t^{2n})|_0^x $
$= -x+\frac{1}{2}x^2-...+\frac{1}{2n}x^{2n}=\sum_{k=1}^{2n}\frac{(-1)^{k}x^k}{k}$ done. 
A: Expand $\frac{1}{1+t}$ as a geometric series. This is different from the series for the logarithm.
