Find Maclaurin series and radius of convergence for $f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2}$ Given $f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2}$, find  the Maclaurin series and radius of convergence.
My attempt:
By considerably simple computation we get that
$f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2} = \ln(1+x) -\frac{1}{(x+1)}$, for $t \ge 0$,
Now what I'm confused about is that the Maclaurin series of $\displaystyle \ln(x+1)=\sum_{n=o}^\infty (-1)^{n+1} \frac{x^n}{n}$, and $\frac{1}{(t+1)} = \left( \ln(t+1) \right) ' $
does that means that: $\displaystyle f(x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} - (\frac{1}{x+1})$ but it seems to me that the convergence radius is $R=1$, but I'm missing the algebraic step to achieve that

$$\int_0^x \frac{t}{(t+1)^2} \, dt=\left[\frac{1}{t+1}+\log (t+1)\right]_0^x=\log (x+1)-\frac{x}{x+1}$$
$$\log(x+1)=\sum _{n=1}^{\infty } \frac{(-1)^{n+1} x^n}{n}$$
$$\frac{x}{x+1}=\sum_{n=1}^{\infty}(-1)^{n+1}x^n$$
$$\int_0^x \frac{t}{(t+1)^2} \, dt=\sum _{n=1}^{\infty } \frac{(-1)^n (n-1) x^n}{n}$$
radius of convergence is $r=1$.
