Sum of $\frac{C_r}{(r+1)(r+2)}$ from $r=0$ to $r=n$ I am trying to compute the following sum: $$\sum_{r=0}^{n}\frac{C_r}{(r+1)(r+2)}$$
Integrating $(1+x)^n = \sum_{r=0}^{n}C_rx^r$ with respect to $x$,
$$\int(1+x)^n dx=\int\sum_{r=0}^{n}C_rx^rdx$$
$$\implies \frac{(1+x)^{n+1}}{n+1}=\sum_{r=0}^{n}C_r\frac{x^{r+1}}{r+1}$$
Again integrating with respect to $x$ from $0$ to $1$ yields,
$$\int_0^1\frac{(1+x)^{n+1}}{n+1}dx=\int_0^1\sum_{r=0}^{n}C_r\frac{x^{r+1}}{r+1}dx$$
$$\implies\bigg[\frac{(1+x)^{n+2}}{(n+1)(n+2)}\bigg]_0^1=\bigg[\sum_{r=0}^{n}C_r\frac{x^{r+2}}{(r+1)(r+2)}\bigg]_0^1$$
$$\implies\sum_{r=0}^{n}\frac{C_r}{(r+1)(r+2)}=\frac{2^{n+2}-1}{(n+1)(n+2)}$$
But it is written in my textbook that the answer is $$\frac{2^{n+2}-(n+3)}{(n+1)(n+2)}$$
So, where am I going wrong? Is it correct to use integral twice in such situations?

We have to respect the integration constant when evaluating an indefinite integral.

When calculating $\int(1+x)^n\,dx = \int\sum_{r=0}^nC_rx^r\,dx$ we obtain
  \begin{align*}
\int(1+x)^n\,dx&=\frac{(1+x)^{n+1}}{n+1}+K_1\\
\int\sum_{r=0}^nC_rx^r\,dx&=\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}+K_2
\end{align*}
  with $K_1,K_2$ constants.

It follows
\begin{align*}
\frac{(1+x)^{n+1}}{n+1}=\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}\color{blue}{+K}\tag{1}
\end{align*}
with $K=K_2-K_1$.
By setting $x=0$ in (1) we obtain
\begin{align*}
\color{blue}{\frac{1}{n+1}=K}
\end{align*}

We conclude
  \begin{align*}
\frac{(1+x)^{n+1}}{n+1}=\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}\color{blue}{+\frac{1}{n+1}}\tag{2}
\end{align*}
Integrating (2) from $0$ to $1$ gives now the correct result
\begin{align*}
\color{blue}{\int_{0}^1\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}\,dx}&=\int_{0}^1\left(\frac{(1+x)^{n+1}}{n+1}-\frac{1}{n+1}\right)\,dx\\
&=\left[\frac{(1+x)^{n+2}}{(n+1)(n+2)}-\frac{x}{n+1}\right]_0^1\\
&=\left[\frac{2^{n+2}}{(n+1)(n+2)}-\frac{1}{n+1}\right]-\left[\frac{1}{(n+1)(n+2)}\right]\\
&\color{blue}{=\frac{2^{n+2}-(n+3)}{(n+1)(n+2)}}
\end{align*}
  as expected.

