The value of the integral $\int_{\pi / 6}^{\pi / 2} \left( \frac{1+\sin 2x+\cos 2x}{\sin x+\cos x} \right) dx$ is equal to

$\int_1^e \frac{1}{x} \, dx$ is equal to

$\int_3^8 \frac{2 - 3x}{x\sqrt{1 + x}} \, dx$ is equal to

Let $f(x) = 2 + |x| - |x - 1| + |x + 1|$,$x \in R$. Consider:
$(S1): f^{\prime}\left(-\frac{3}{2}\right) + f^{\prime}\left(-\frac{1}{2}\right) + f^{\prime}\left(\frac{1}{2}\right) + f^{\prime}\left(\frac{3}{2}\right) = 4$
$(S2): \int_{-2}^{2} f(x) dx = 12$
Then,

For $0 < x < \frac{\pi}{2}$,the integral $\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \ln(e^{\cos x}) \, d(\sin x)$ is equal to:

$\int_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\,=$