summation - Evaluate $sum_{j=0}^n(-1)^{n+j}{nchoose j}{{n+j}choose j}frac{1}{(j+1)^2}$

I want to evaluate the sum

$\displaystyle \sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{(j+1)^2}$

My approach so far has been the possible use of shifted Legendre polynomial. We know

$Q_n(x)=\displaystyle \sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}x^j$

But I am not being able to relate these two. I see that

$\displaystyle \int_0^1Q_n(x)dx=\sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{j+1}$

but how do I get $\dfrac{1}{(j+1)^2}$ ? I need some hint for this.

Answer

Use Kelenner's hint about
$$\int_{0}^{1}x^j(-\log x)\,dx = \frac{1}{(1+j)^2}\tag{1}$$
and exploit the fact that $(-\log x)$ has a nice representation in terms of shifted Legendre polynomials:

$$(-\log x) = 1+\sum_{j\geq 1}\frac{(-1)^j(2j+1)}{j(j+1)}Q_j(x)\tag{2}$$
since, for any $n\geq 1$,
$$\int_{0}^{1}(-\log x)Q_n(x)\,dx = \color{red}{\frac{(-1)^n}{n(n+1)}}\tag{3}$$
can be proved through Rodrigues' formula and integration by parts (the derivative of $(-\log x)$ is simple to deal with).