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
 A: $$\int_0^x \frac{t}{(t+1)^2} \, dt=\left[\frac{1}{t+1}+\log (t+1)\right]_0^x=\log (x+1)-\frac{x}{x+1}$$
$$\log(x+1)=\sum _{n=1}^{\infty } \frac{(-1)^{n+1} x^n}{n}$$
$$\frac{x}{x+1}=\sum_{n=1}^{\infty}(-1)^{n+1}x^n$$
$$\int_0^x \frac{t}{(t+1)^2} \, dt=\sum _{n=1}^{\infty } \frac{(-1)^n (n-1) x^n}{n}$$
radius of convergence is $r=1$.
A: You don't have to explicitly compute the integral, nor the successive derivatives: by the 1st fundamental theorem of integral calculus, $f(x)$ is the function such that
$$f'(x)=\frac x{(1+x)^2},\qquad f(0)=0.$$
Therefore, we only have to find the expansion of $\dfrac x{(1+x)^2}$ and integrate term by term. The radius of convergence will be the same for $f(x)$ and $f'(x)$.
Now, $\dfrac 1{(1+x)^2}=-\Bigl(\dfrac1{1+x}\Bigr)'=\displaystyle\biggl(\sum_{n\ge0}(-1)^{n+1} x^n\biggr)'$, so
$$f'(x)=x\sum_{n\ge0}(-1)^{n+1}n\, x^{n-1}=\sum_{n\ge0}(-1)^{n+1}n\, x^n,$$
whence
$$f(x)=\sum_{n\ge0}(-1)^{n+1}\frac{n\,x^{n+1}}{n+1}.$$
A: There is no need to integrate. $f(0)=0$, obviously, and
$$f'(x)=\frac x{(1+x)^2}=\frac1{1+x}-\frac1{(1+x)^2}.$$
From this,
$$f^{(n+1)}(0)=(-1)^nn!+(-1)^{n+1}(n+1)!$$
So the general term of Taylor is
$$(-1)^n\left(\dfrac1{n+1}-1\right)x^{n+1}$$ and by the ratio test, this converges for $|x|<1$.
