Using Variation of Parameters to Find the General Solution of $u''-u=\frac{2}{e^x+1}$ 
I am trying to use variation of parameters to find the general solution to the inhomogeneous ODE
  $$u''-u=\frac{2}{e^x+1}$$

By computing the characteristic polynomial of the homogeneous equation $u''-u=0$, I have found $$u_H=c_1e^x+c_2e^{-x} \ \ \ \ c_1,c_2\in\mathbb{R}$$
Hence $u_1(x)=e^x, u_2(x)=e^{-x}$. So,
\begin{align}
v'_1(x)&=-\frac{u_2(x)f(x)}{W(x)} \\
v_1(x)&=\int \frac{e^{-x}}{e^x+1} \ dx \\
&=-e^{-x}+\ln|e^{-x}+1|+C_1
\end{align}
and similarly, 
\begin{align}
v'_2(x)&=\frac{u_1(x)f(x)}{W(x)} \\
v_2(x)&=-\int \frac{e^{x}}{e^x+1} \ dx \\
&=-\ln|e^{x}+1|+C_2
\end{align}
Hence the general solution is given by
$$u(x)=v_1(x)u_1(x)+v_2(x)u_2(x)=\left(-e^{-x}+\ln|e^{-x}+1|+C_1\right)e^x+\left(-\ln|e^{x}+1|+C_2\right)e^{-x}$$
But this is incorret. Is my method valid?
 A: First, 
$$W(x)=\det\left(\begin{bmatrix}\exp(x)&\exp(-x)\\\exp(x)&-\exp(-x)\end{bmatrix}\right)=-2\,.$$
That is, with a proper constant of integration,
$$\begin{align}v_1(x)&=-\int\,\frac{u_2(x)\,f(x)}{W(x)}\,\text{d}x
\\&=-\exp(-x)+\ln\big(1+\exp(-x)\big)=-\exp(-x)-x+\ln\big(1+\exp(x)\big)\,.\end{align}$$
Likewise, for an appropriate choice of the constant of integration, we have
$$v_2(x)=+\int\,\frac{u_1(x)\,f(x)}{W(x)}\,\text{d}x=-\ln\big(1+\exp(x)\big)\,.$$
Thus, a particular solution is
$$\begin{align}u_p(x)&=v_1(x)\,u_1(x)+v_2(x)\,u_2(x)
\\&=\small\exp(+x)\,\Big(-\exp(-x)-x+\ln\big(1+\exp(x)\big)\Big)-\exp(-x)\,\ln\big(1+\exp(x)\big)
\\
&=-1-x\,\exp(x)+2\,\sinh(x)\,\ln\big(1+\exp(x)\big)\,.\end{align}$$
Hence, the general solutions are of the form
$$u(x)=u_p(x)+a\,\exp(+x)+b\,\exp(-x)\,,$$
where $a$ and $b$ are constants.  My result agrees with you.

Alternatively, note that
$$\frac{\text{d}}{\text{d}x}\,\exp(x)\,\big(u'(x)-u(x)\big)=\frac{2\,\exp(x)}{\exp(x)+1}\,.$$
That is,
$$u'(x)-u(x)=\int\,\frac{\exp(x)}{\exp(x)+1}\,\text{d}x=2\,\exp(-x)\,\ln\big(\exp(x)+1\big)-2A\,\exp(-x)$$ 
for some constant $A$.  That is,
$$\frac{\text{d}}{\text{d}x}\,\exp(-x)\,u(x)=2\,\exp(-2x)\,\ln\big(\exp(x)+1\big)-2A\,\exp(-2x)\,.$$
Ergo,
$$u(x)=\exp(+x)\,\int\,2\,\exp(-2x)\,\ln\big(\exp(x)+1\big)\,\text{d}x+A\,\exp(-x)\,.$$
Using integration by parts, we obtain
$$\int\,2\,\exp(-2x)\,\ln\big(\exp(x)+1\big)\,\text{d}x=-\exp(-x)-x+2\exp(-x)\,\sinh(x)\,\ln\big(1+\exp(x)\big)+B$$
for some constant $B$, and so we conclude that
$$u(x)=-1-x\,\exp(x)+2\,\sinh(x)\,\ln\big(1+\exp(x)\big)+A\,\exp(-x)+B\,\exp(+x)\,.$$
A: For particular solution, we can use differential operator method for $D(u''-u)=\dfrac{2}{e^x+1}$ then
$$(D^2-1)u=\frac{2}{e^x+1}$$
$$u=\dfrac{1}{D-1}\left(\dfrac{1}{D+1}(\frac{2}{e^x+1})\right)$$
let $v=\dfrac{1}{D+1}(\dfrac{2}{e^x+1})$ then
$$e^xv'+e^xv=\dfrac{2e^x}{e^x+1}$$
gives the function $v=e^{-x}\int\dfrac{2e^x}{e^x+1}dx=2e^{-x}\ln(e^x+1)$. Now 
$$u=\dfrac{1}{D-1}\left(2e^{-x}\ln(e^x+1)\right)$$
or
$$e^{-x}u'-e^{-x}u=2e^{-2x}\ln(e^x+1)$$
therefore 
$$u=e^{x}\int2e^{-2x}\ln(e^x+1)dx=\color{blue}{e^x\ln\left(e^{-x}+1\right)-e^{-x}\ln \left(e^x+1\right)-1}$$
A: $$u''-u=\frac{2}{e^x+1}$$
$$u''-u'+u'-u=\frac{2}{e^x+1}$$
Substitute $s=u'-u$
$$s'+s=\frac{2}{e^x+1}$$
$$(se^x)'=\frac{2e^x}{e^x+1}$$
Integrating
$$se^x=2\ln({e^x+1})+K_1$$
$$(u'-u)e^{-x}=2e^{-2x}\ln({e^x+1})+K_1e^{-2x}$$
$$(ue^{-x})'=2e^{-2x}\ln({e^x+1})+K_1e^{-2x}$$
After integration
$$u=2e^x\int \ln({e^x+1})e^{-2x}dx+K_1e^{-x}+K_2e^x$$
$$u=-\ln({e^x+1})e^{-x}+e^x\int \frac {e^{-x}}{e^x+1}dx+K_1e^{-x}+K_2e^x$$
Finally
$$u(x)=-\ln({e^x+1})e^{-x}+e^x\ln( {e^{-x}+1})-1+K_1e^{-x}+K_2e^x$$
or
$$u(x)=\ln({e^x+1})(-e^{-x}+e^x)-1-xe^x+K_1e^{-x}+K_2e^x$$

Edit
I think the $−xe^x$ is wrong
It's not wrong I think it comes from the
$$e^x\ln(e^{-x}+1)=e^x\ln (e^x+1)-xe^x$$
