Please help to continue the derivation of closed form of zeta function at odd integer values

Question

There would be many questions in stack exchange asking about the value at odd integer values, but in this question, I have done my own work but stuck at a point. I don't know many properties of the zeta function that would come to use here. Also, tell me if I am going in the wrong direction.
I had seen the well known formula(of Euler) that gives the value of $\zeta(2n)$, and I wondered if there was a formula for odd integers. I tried it today. I will skip some steps like integrals which can be evaluated easily. Here is my work:
Let $x$ and $y$ be real numbers with real part greater than or equal to $1$. So \begin{align} \begin{split} \zeta(x+y)={}&\sum_{k\geq 1}\frac{1}{k^{x}k^{y}}\\ ={}&\sum_{k\geq 1}\frac{1}{y+1}\int_{0}^{1}(x+r)k^{x+r-2}dr\\ ={}&\sum_{k\geq 1}\int_{0}^{1}(x+r)k^{x+r+y-2}dr\\ ={}&\sum_{k\geq 1}(x+r)\int_{0}^{1}k^{x+r+y-2}dr\\ ={}&\int_{0}^{1}(x+r)\zeta(2-x-y-r)dr\\ \end{split} \end{align} Now let $x$ be an integer.
Substituting $2x$ instead of $x$ and $1$ instead of $y$ gives \begin{align} \begin{split} \zeta(2x+1)={}&\int_{0}^{1}(2x+r)\zeta(1-2x-r)dr\\ ={}&(x^{2}+rx)|_{r=0}^{r=1}-\int_{0}^{1}(x^{2}+rx)\zeta'(1-2x-r)dr\\ \end{split} \end{align} We know that $$\zeta'(s)=\sum_{k\geq 1}\frac{-\log k}{k^s}$$ So \begin{align} \begin{split} (x^{2}+rx)|_{r=0}^{r=1}-\int_{0}^{1}(x^{2}+rx)\zeta'(1-2x-r)dr={}&x+\int_{0}^{1}\sum_{k\geq 1}\frac{\log k}{k^{1-2x-r}}(x^{2}+rx)dr\\ ={}&x+\sum_{k\geq 1}\log k\int_{0}^{1}\frac{x^{2}+rx}{k^{1-2x-r}}dr\\ ={}&x+\sum_{k\geq 1}\frac{xk^{2x-1}((kx+k-x)\log(k)-k+1)}{\log k} \end{split} \end{align} Now I don't know what to do. I don't think that there is a closed form of the sum. If there isn't any closed form of the sum, I would be satisfied with any value of the sum. Also, I want to ask did I make any mistake?
Update:I expanded the sum and it became more complicated: $$(x+1)\zeta(-2x)-x\zeta(1-2x)-\frac{\zeta(-2x)}{2}+\frac{1}{2}\sum_{k\geq 1}\frac{k^{2x}(2x^{2}\log^{2}k-2x\log k+1)}{4x^{3}\log^{2}k}+\sum_{k\geq 1}\frac{1}{\log k}$$ Another question arises from here: I we expand the last sum(the big one, not the $\frac{1}{\log k}$ sum), a similar sum(added to some other addends related to the zeta function) comes out. What is the formula for the $n$th sum that comes out? I am not asking for the value of the $n$th sum, I am asking for the formula for $a_k$ where the $n$ sum is $\sum_{k\geq 1}a_k$.