Involving Harmonic number 
$$\int_{0}^{\infty}{x^n\ln{x}\over (\sinh{x}+\cosh{x})^2}\mathrm dx={n!\over 2^{n+1}}(H_n-\gamma-\ln{2})\tag1$$

$H_n$ is the nth-Harmonic number, where $H_0=0$, $\gamma$ is the Euler-Mascheroni constant.

How does one show that $(1)={n!\over 2^{n+1}}(H_n-\gamma-\ln{2})?$

$$(\sinh{x}+\cosh{x})^2=\sinh^2{x}+\cosh^2{x}+\sinh{(2x)}$$
$$=2\sinh^2{x}+\sinh(2x)+1$$
$$\int_{0}^{\infty}{x^n\ln{x}\over 2\sinh^2{x}+\sinh(2x)+1}\mathrm dx\tag2$$
$x=e^y$ then $\mathrm dx=e^y\mathrm dy$
$$\int_{-\infty}^{\infty}{ye^{y(n+1)}\over 2\sinh^2{e^y}+\sinh{(2e^y)}+1}\mathrm dy\tag3$$
$\sinh(x)={1\over 2}(e^x-e^{-x})$
$\sinh^2(x)={1\over 2}(e^{2x}+e^{-2x}-2)$
OR
$$\int_{0}^{\infty}{x^n\ln{x}\over (\sinh{x}+\cosh{x})^2}\mathrm dx=\int_{0}^{\infty}{x^n\ln{x}\over e^{(2x)}}\mathrm dx\tag4$$
$(4)$ probably apply IBP...
 A: $$\int_{0}^{+\infty}x^\alpha \log(x) e^{-2x}\,dx = \frac{d}{d\alpha}\int_{0}^{+\infty}x^{\alpha} e^{-2x}\,dx \tag{A}$$
and since
$$ \int_{0}^{+\infty}x^\alpha e^{-2x}\,dx = \frac{\Gamma(\alpha+1)}{2^{\alpha+1}} \tag{B}$$
and $\frac{d}{d\alpha} f = f\cdot \frac{d}{d\alpha}\log f$ we have
$$ \int_{0}^{+\infty}x^\alpha \log(x)e^{-2x}\,dx = \frac{\Gamma(\alpha+1)}{2^{\alpha+1}}\left[\psi(\alpha+1)-\log 2\right]\tag{C}$$
and the claim readily follows from the known relation between $\psi(n+1)$ and $H_n$.
A: Observing that $\sinh x + \cosh x = e^x$, the integral can be rewritten as
$$I = \int^\infty_0 e^{-2x} x^n \ln x \, dx.$$
Now consider the integral
$$I_\alpha = \int^\infty_0 e^{-2x} x^n \, dx.$$
Setting $x \mapsto 2x$ we find
$$I_\alpha = \frac{1}{2^{n + 1}} \Gamma (n + 1).$$
Writing 
$$\Gamma (n + 1) = 2^{n + 1} I_\alpha = 2^{n + 1} \int^\infty_0 e^{-2x} x^n \, dx,$$
differentiating with respect to the parameter $n$ we have
\begin{align*}
\frac{d}{dn} \Gamma (n + 1) &= \frac{d}{dn} \left \{2^{n + 1} \int^\infty_0 e^{-2x} x^n \, dx \right \}\\
\Gamma' (n + 1) &= \ln 2 \cdot 2^{n + 1} \int^\infty_0 e^{-2x} x^n \, dx + 2^{n + 1} \int^\infty_0 e^{-2x} x^n \ln x \, dx\\
&= \ln 2 \cdot 2^{n + 1} I_\alpha + 2^{n + 1} I,
\end{align*}
or
$$I = \frac{1}{2^{n + 1}} \left (\Gamma' (n + 1) - \ln 2 \cdot \Gamma (n + 1) \right ).$$
As
$$\Gamma' (n + 1) = \psi (n + 1) \Gamma (n + 1),$$
where $\psi (x)$ is the digamma function, we can write
$$I = \frac{\Gamma (n + 1)}{2^{n + 1}} \left \{\psi (n + 1) - \ln 2 \right \}.$$
Also, from the following well-known property for the digamma function of
$$\psi (n) = H_{n -1} - \gamma,$$
one has
$$\int^\infty_0 \frac{x^n \ln x}{(\sinh x + \cosh x)^2} \, dx = \frac{\Gamma (n + 1)}{2^{n + 1}} \left (H_n - \gamma - \ln 2 \right ),$$
as required to show.  
