Integrating to derive relation between binomial co-efficient 
If $C_o,C_1...$ are the binomial coefficents in the expansion of $(1+x)^n$ and $$\sum_{r=0}^{n} (-1)^r \binom{n}{r} \frac{1}{(r+1)^2}  = k \sum_{r=0}^{n} \frac{1}{r+1}$$
Find a k such that the above equation is satisfied

My attempt:
$$ (1+x)^n = \sum_{k=0}^{k=n} \binom{n}{k} x^k$$
integrate both sides
$$ \frac{ (1+x)^{n+1} }{n+1}  =\sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{k+1} +C$$
$$ x= 0 $$
$$  \implies \frac{-n}{n+1}  = C$$
Divide both sides by 'x' and integrate
$$ \int \frac{ (1+x)^{n+1} }{  (x) n+1}  dx = \sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{(k+1)^2} - \frac{n}{n+1} \ln(x)+ C'$$
Or,
$$ \int \frac{ (1+x)^{n+1} }{  (x) n+1}  dx  - \frac{n}{n+1} \ln(x)+ C' = \sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{(k+1)^2}$$
Not so sure what do here, like what to put as bounds. Slightly concerned I may have to evaluate a negative logarithm
 A: Here is my approach. Let denote that $H_n = \sum_{k=1}^{n}\frac{1}{k}$ is the n-th Harmonic series. Then it is easy to observe that:
$$H_n = \sum_{k=1}^{n}\frac{1}{k} = \sum_{k=1}^{n}\int_{0}^{1} x^{k-1}dx = \int_{0}^{1}\sum_{k=1}^{n} x^{k-1}dx = \int_{0}^{1} \frac{1-x^n}{1-x}~dx$$
For your approach above, instead of using $(1+x)^n$, why don't you use $(1-x)^n$?
$$(1-x)^n = \sum_{k=0}^{n} \binom{n}{k}(-1)^r x^r \Rightarrow \int_{0}^{t} (1-x)^n ~dx = \int_{0}^{t}\sum_{k=0}^{n} \binom{n}{k}(-1)^r x^r dx$$$$\Rightarrow \frac{1- (1-t)^{n+1}}{t(n+1)}= \sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^r t^{r}}{r+1}$$
Now, notice that:
$$LHS=\sum_{r=0}^{n} \binom{n}{r}\frac{(-1)^r}{(r+1)^2}=\sum_{r=0}^{n} \binom{n}{r}\frac{(-1)^rt^{r+1}}{(r+1)^2}\Bigg|_0^1=\int_{0}^{1}\sum_{r=0}^{n} \binom{n}{r}\frac{(-1)^r t^{r}}{r+1}~dt$$$$=\int_{0}^{1} \frac{1- (1-t)^{n+1}}{t(n+1)} ~ dt$$
Here we will move to the RHS:
$$RHS=k \sum_{r=0}^{n} \frac{1}{r+1} = k\cdot H_{n+1} = k \cdot \int_{0}^{1} \frac{1-t^{n+1}}{1-t}~dt= k \int_{0}^{1} \frac{1-(1-t)^{n+1}}{t}~dt \cdot $$
By comparing $LHS$ with $RHS$, we obtain that $k=\frac{1}{n+1}$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{r = 0}^{n}\pars{-1}^{r}{n \choose r}
{1 \over \pars{r + 1}^{2}}} =
\sum_{r = 0}^{n}\pars{-1}^{r}{n \choose r}\
\overbrace{\bracks{-\int_{0}^{1}\ln\pars{x}\, x^{r}\,\dd x}}
^{\ds{1 \over \pars{r + 1}^{2}}}
\\[5mm] = &\
-\int_{0}^{1}\ln\pars{x}\sum_{r = 0}^{n}{n \choose r}\pars{-x}^{r}\,\dd x =
-\int_{0}^{1}\ln\pars{x}\pars{1 - x}^{n}\,\dd x
\\[5mm] = &\
 -\bracks{\xi^{1}}\int_{0}^{1}x^{\xi}\pars{1 - x}^{n}\,\dd x =
 -\bracks{\xi^{1}}\bracks{\Gamma\pars{\xi + 1}\Gamma\pars{n + 1} \over \Gamma\pars{\xi + n + 2}}
\\[5mm] = &\
 -n!\bracks{\xi^{1}}\bracks{%
\Gamma\pars{1} + \Gamma\, '\pars{1}\xi \over \Gamma\pars{n + 2} +
\Gamma\, '\pars{n + 2}\xi}
\\[5mm] = &\
-\,{1 \over n + 1}\bracks{\xi^{1}}\bracks{%
1 - \gamma\xi \over 1 + \Psi\pars{n + 2}\xi}
\\[5mm] = &\
-\,{1 \over n + 1}\bracks{\xi^{1}}\braces{\vphantom{\Large A}%
\pars{\vphantom{\large A}1 - \gamma\xi}\bracks{\vphantom{\large A}1 - \Psi\pars{n + 2}\xi}}
\\[5mm] = &\
-\,{1 \over n + 1}\bracks{-\gamma - \Psi\pars{n + 2}} =
-\,{1 \over n + 1}\pars{-H_{n + 1}} =
\\[5mm] = &\
{1 \over n + 1}\sum_{r = 0}^{n}{1 \over r + 1}
\implies \bbx{k = {1 \over n + 1}} \\ &
\end{align}

$\ds{\bracks{\xi^{m}}}$ is the
Coefficient Extraction Operator.

$\ds{\Gamma}$ is the
Gamma Function.

$\ds{\gamma}$
is the Euler-Mascheroni Constant.

$\ds{\Psi}$ is the
Digamma Function.

$\ds{H_{z}}$ is a
Harmonic Number.
A: In trying to evaluate
$$S_n = \sum_{r=0}^n (-1)^r {n\choose r} \frac{1}{(r+1)^2}$$
we introduce
$$f(z) = \frac{(-1)^n \times n!}{(z+1)^2} 
\prod_{q=0}^n \frac{1}{z-q}$$
which has the property that for $0\le r\le n$
$$\mathrm{Res}_{z=r} f(z) =
\frac{(-1)^n \times n!}{(r+1)^2}
\prod_{q=0}^{r-1} \frac{1}{r-q}
\prod_{q=r+1}^n \frac{1}{r-q}
\\ = \frac{(-1)^n \times n!}{(r+1)^2}
\frac{1}{r!} \frac{(-1)^{n-r}}{(n-r)!}
= (-1)^r {n\choose r} \frac{1}{(r+1)^2}.$$
It follows that
$$S_n = \sum_{r=0}^n \mathrm{Res}_{z=r} f(z).$$
Now residues sum to zero and the residue at infinity of $f(z)$ is zero
by inspection, therefore
$$S_n = - \mathrm{Res}_{z=-1} f(z) =
(-1)^{n+1} \times n! \times
\left. \left( \prod_{q=0}^n \frac{1}{z-q} \right)' \right|_{z=-1}
\\ = (-1)^{n+1} \times n! \times
\left. \prod_{q=0}^n \frac{1}{z-q} 
\sum_{q=0}^n \frac{1}{q-z} \right|_{z=-1}
\\ = (-1)^{n+1} \times n! \times
(-1)^{n+1} \prod_{q=0}^n \frac{1}{q+1}
\times \sum_{q=0}^n \frac{1}{q+1}
\\ = n! \times \frac{1}{(n+1)!}
\times \sum_{q=0}^n \frac{1}{q+1}
= \frac{1}{n+1} \sum_{q=0}^n \frac{1}{q+1} = \frac{1}{n+1} H_{n+1}.$$
We see that the desired factor on the sum is $k=\frac{1}{n+1}.$
