Is it possible to get a closed-form for $\int_0^1\frac{(1-x)^{n+2k-2}}{(1+x)^{2k-1}}dx$? It is know that
$$H_n=-n\int_0^1(1-t)^{n-1}\log (t)dt,$$
see [1], where $H_n=1+1/2+\ldots+1/n$ it the nth harmonic number. Then I believe that can be used for $x>0$ 
$$\frac{1}{2}\log(x)=\sum_{k=1}^{\infty}\frac{(-1)^{2k-1}}{2k-1}\left(\frac{x-1}{1+x}\right)^{2k-1},$$
see for example [2], to show 
$$H_n=2n\sum_{k=1}^{\infty}\frac{1}{2k-1}\int_0^1\frac{(1-x)^{n+2k-2}}{(1+x)^{2k-1}}dx.$$

Question. It is possible to get a closed-form for 
  $$\int_0^1\frac{(1-x)^{n+2k-2}}{(1+x)^{2k-1}}dx?$$

If you want give a justification for previous expression for $H_n$, and know how comoute previous definite integral I am wait your answer. Thanks in advance. 
References:
[1] Furdui, LA GACETA de la Real Sociedad Matemática Española. Third paragraph in page 699, see here.
[2] Hyslop, Infinite Series, Dover Publications (2006).

(More of a comment, since I think you suggest you know this, but maybe this is useful to somebody else answering.)
According to Mathematica if $k,n\in\mathbb{Z}$ and $2 k+n>1$,
$$I=\frac{\, _2F_1(1,2 k-1;2 k+n;-1)}{2 k+n-1}$$
Where $F$ is the hypergeometric function on MathWorld.
