The U-sub isn't completely obvious

Relevant wiki: Integration Tricks

Use $\small{\color{teal}{1-\cos x=2\sin^2\frac x2}}$ and $\small{\color{teal}{\sin x=2\sin \frac x2\cos \frac x2}}$ and cancel factor of $\small{2\sin \frac x2}$ from both numerator and denominator.

$\large\mathscr E=\displaystyle\int_0^{\pi/4}\dfrac{\cos \frac x2-\sin \frac x2}{\cos \frac x2+\sin \frac x2}\mathrm{d}x$

Substitute $\cos \frac x2+\sin \frac x2=t$ so that integral is of the form $\displaystyle\int \dfrac{2\mathrm{d}t}{t}=2\ln t+C$ $\large =2\left[\ln\left(\cos \frac x2+\sin \frac x2\right)\right]_0^{\pi/4}$ $\large =\left[\ln\left(\cos \frac x2+\sin \frac x2\right)^2\right]_0^{\pi/4}$ $\large =\left[\ln\left(1+\sin x\right)\right]_0^{\pi/4}$ $\large =\ln\left(1+\dfrac1{\sqrt 2}\right)$

$\Large \therefore 1+1+2=\boxed 4$

$\mathscr{E} = \displaystyle \int_{0}^{\pi/4} \dfrac{\sin x +\cos x-1}{\sin x - \cos x + 1} dx\\ =\displaystyle \int_{0}^{\pi/4} \dfrac{\left(\sin x +\cos x-1\right)\color{#3D99F6}{\left(\cos x - \sin x + 1\right)}}{\left(\sin x - \cos x + 1\right)\color{#3D99F6}{\left(\cos x - \sin x + 1\right)}}dx\\ =\displaystyle \int_{0}^{\pi/4} \dfrac{\color{#D61F06}{\sin x \cos x} - \sin^2 x \color{#EC7300}{+ \sin x} \color{#69047E}{+ \cos^2 x}\color{#D61F06}{-\sin x \cos x}\color{#20A900}{+\cos x- \cos x}\color{#EC7300}{+\sin x} - 1}{\color{#D61F06}{\sin x \cos x} - \sin^2 x\color{#EC7300}{+\sin x}-\cos^2 x \color{#D61F06}{+\sin x \cos x}\color{#20A900}{ - \cos x +\cos x} \color{#EC7300}{- \sin x} + 1}dx\\ =\displaystyle \int_{0}^{\pi/4} \dfrac{2 \sin x - \sin^2 x - 1 \color{#69047E}{+1-\sin^2 x}}{2\sin x \cos x - \color{#69047E}{(\cos^2 x + \sin^2 x)}+1}dx\\ =\displaystyle \int_{0}^{\pi/4} \dfrac{2 \sin x - 2\sin^2 x}{2\sin x \cos x}dx\\ =\displaystyle \int_{0}^{\pi/4} \left(\dfrac{1}{\cos x} +\dfrac{-\sin x}{\cos x} \right)dx\\ =\displaystyle \int_{0}^{\pi/4} \left(\dfrac{\sec x(\sec x+ \tan x)}{\sec x+ \tan x} +\dfrac{-\sin x}{\cos x} \right)dx\\ =\displaystyle \int_{0}^{\pi/4} \left(\dfrac{\sec x\tan x + \sec^2 x}{\sec x+ \tan x} +\dfrac{-\sin x}{\cos x} \right)dx\\ =\Big[\ln\left(\sec x+\tan x\right)+\ln\left(\cos x\right)\Big]_{0}^{\pi/4}\\ =\Big[\ln\left(\cos x\left(\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}\right)\right)\Big]_{0}^{\pi/4}\\ =\Big[\ln\left(1+\sin x\right)\Big]_{0}^{\pi/4}\\ =\ln \left(1+\sin \dfrac{\pi}{4}\right) - \ln\left(1+\sin 0\right)\\ =\ln \left(1+\dfrac{1}{\sqrt{2}}\right) - \ln(1+0)\\ =\ln \left(1+\dfrac{1}{\sqrt{2}}\right)$

$\implies a =1,\;b=1,\;c=2,\;a+b+c=1+1+2=\boxed{4}$