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

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 $k=\frac{1}{n+1}$
