On Abel summation $\sum_{e<n\leq x}\left(\mu(n)\cdot\int_2^n\frac{ds}{\log s}\right)$, where $\mu(n)$ is the Möbius function

By Abel's identity for $Li (x)=\int_2^x\frac{ds}{\log s}$, $a(n)=\mu(n)$ the Möbius function and $[y=e,x]$ (see Theorem 4.2, page 77 of [1]) and an application of Fundamental Calculus Theorem we obtain

$$\sum_{e <n\leq x}\mu (n)Li (n)=\sum_{e<n\leq x}\mu(n)\cdot\left(\int_2^n\frac{ds}{\log s}\right)=M(x)\cdot\int_2^x\frac{dt}{\log t}-\int_e^x\frac{M(t)}{\log t}dt,$$ where $M(x)=\sum_{1\leq n\leq x}\mu(n)$. Since we have the obvious estimating $M(x)=O(x)$ we have $$\int_e^x\frac{M(t)}{\log t}dt=O\left(\int_e^x\frac{t}{\log t}dt\right)$$ I believe that $\int_e^x (t/\log t) dt$ is computed using definition of exponential integral, by I don't know how made it (by change of variables we obtain $\int_1^{\log x}(e^{2t}/t) dt$ ).

My

Question. a) A proof verification of previous computations is required and will be appreciated. b) How bound in a more accurate way, if it is possible, $\int_e^x M(t)/\log t dt$? c) How bound $\int_e^x t/\log t dt$ (I search an answer to c), regardless of your answer to b) I say a big oh- bound too)?

Thanks in advance. My goal is strengthen my standard maths and encourage people.

References:

[1] Apostol, Introduction to analytitc Number Theory, Springer 1978.

a) Your computations are correct, the only thing I note that the function $\int_{2}^{n}dt/\log\left(t\right)$ has a name; it's the logaritmic integral, and the standard notation is $\textrm{Li}\left(n\right)$.
b-c) Assuming Riemann hypothesis, we can estimate the Merten's function as $M\left(x\right)=O\left(x^{1/2+\epsilon}\right),\,\forall\epsilon>0$. Actually there are some improvements of this bound, for more details see here. About the integrals, the only “closed form” you can obtain involving the Exponential integral function, but if you want a “more tractable” form we can use the well-knonw asymptotic $$\textrm{Li}\left(x\right)\sim\frac{x}{\log\left(x\right)}$$ hence, assuming RH, $$\int_{e}^{x}\frac{M\left(t\right)}{\log\left(t\right)}dt=O\left(\frac{x^{3/2+\epsilon}}{\log\left(x\right)}\right),\,\forall\epsilon>0$$ and without RH, $$\int_{e}^{x}\frac{M\left(t\right)}{\log\left(t\right)}dt=O\left(\frac{x^{2}}{\log\left(x\right)}\right).$$