That pesky 1

Call the integration $\mathfrak E$ and then apply $\displaystyle\int_a^b f(x)\mathrm{d}x=\displaystyle\int_a^b f(a+b-x)\mathrm{d}x$ and add them to get:- $\begin{aligned}\mathfrak{E}=&\dfrac 12\left(\large \displaystyle \int_{0}^{\pi/2} \dfrac{\color{teal}{\sin x +\cos x}+ 2}{\color{teal}{\sin x + \cos x}} \, \mathrm{d}x\right)\\=&\dfrac{\pi}{4}+\underbrace{\displaystyle\int_0^{\pi/2}\dfrac{\mathrm{d}x}{\sin x+\cos x}}_{\mathfrak R}\\\mathfrak R=&\displaystyle\int_0^{\pi/2}\dfrac{\mathrm{d}x}{\underbrace{\cos\left(\pi/2 -x\right)+\cos x}_{2\cos (\pi/4)\cos (\pi/4-x)}}\\=&(1/\sqrt{2})\left(\displaystyle\int_0^{\pi/2}\sec(\pi/4-x)\ \mathrm{d}x\right)\\=& (-1/\sqrt 2)\left[\ln(\sec (\pi/4-x)+\tan (\pi/4-x))\right]_0^{\pi/2}\\=&(\underbrace{1/\sqrt 2}_{\color{#D61F06}{\cos (\pi/4)}})\ln\left(\left( \dfrac{\sqrt 2+1}{\sqrt 2-1}\right)\right)\end{aligned}$

Hence, combining we get;-

$\large \mathfrak E=\pi/4+\cos\left(\dfrac{\pi}{4}\right)\ln\left(\dfrac{\sqrt 2+1}{\sqrt 2-1}\right)$

$\large\therefore ~4+4+2+1=\Huge\boxed{\color{#20A900}{11}}$