Closed form of $\int_{0}^{\infty}\sum\limits_{k=0}^{n}{{n\choose k}e^{-(2k+1)x}\over \cosh^n(x)}\mathrm dx$

Question

How do I find the closed form of $(1)?$

$$\int_{0}^{\infty}\sum_{k=0}^{n}{{n\choose k}e^{-(2k+1)x}\over \cosh^n(x)}\mathrm dx=F(n)\tag1$$ Where $n\ge1$

F(n) seems only to give rational number according to numerical observation.

I was able to find the closed form of similar to $(1)$

$$\int_{0}^{\infty}\sum_{k=0}^{n}{{n\choose k}e^{-(2k+1)x}\over \cosh(x)}\mathrm dx={2^n-1\over n}\tag2$$

Where $n\ge1$

Others

$$\int_{0}^{\infty}{e^{-x}\over \cosh(x)}\mathrm dx=\ln{2}\tag3$$

$$\int_{0}^{\infty}{2e^{-2x}\over \cosh(x)}\mathrm dx=4-\pi\tag4$$

The idea came from here

Answer 1

Use that

$$(1+x)^n = \sum_{k=0}^n {n \choose k} x^k$$

Hence

$$\int_{0}^{\infty}\sum_{k=0}^{n}{{n\choose k}e^{-(2k+1)x}\over \cosh^n(x)} = \int^\infty_0 \frac{e^{-x}(1+e^{-2x})^n}{\cosh^n(x)}\,dx$$

Now use that

$$\cosh(x) = \frac{1+e^{-2x}}{2e^{-x}}$$

$$\int^\infty_0 \frac{2^ne^{-x(1+n)}(1+e^{-2x})^n}{(1+e^{-2x})^n}\,dx = 2^n\int^\infty_0 e^{-(n+1)x}\,dx = \frac{2^n}{n+1}$$