ЗНАЧЕНИЕ K=1.Помогите решить

Решите задачу:

$1)\; \; \int\limits^8_1\, \frac{\sqrt{x+2}}{x}\, dx=\Big [\, x+2=t^2\; ,\; t=\sqrt{x+2}\; ,\; x=t^2-2\; ,\; dx=2t\, dt\; \Big ]=\\\\=\Big [\; t_1=\sqrt{1+2}=\sqrt3\; ,\; t_2=\sqrt{8+2}=\sqrt{10}\; \Big ]=\int\limits_{\sqrt3}^{\sqrt{10}}\, \frac{t\cdot 2t\, dt}{t^2-2}=\\\\=2\cdot \int\limits^{\sqrt{10}}_{\sqrt3}\, \frac{t^2\, dt}{t^2-2}\, dt=2\cdot \int\limits^{\sqrt{10}}_{\sqrt3}\, (1+\frac{2}{t^2-2})\, dt=2\cdot (t+\frac{2}{2\sqrt2}\cdot ln\Big |\frac{t-\sqrt2}{t+\sqrt2}\Big |)\Big |_{\sqrt3}^{\sqrt{10}}=$

$=2\cdot \Big (\sqrt{10}-\sqrt3+\frac{1}{\sqrt2}ln\Big |\frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}+\sqrt3}\Big |-\frac{1}{\sqrt2}ln\Big |\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\Big |\Big )$

$2)\; \; \int\limits^{\frac{\pi}{2}}_0\frac{dx}{1+2cosx}=\Big [\, t=tg\frac{x}{2}\; ,\; cosx=\frac{1-t^2}{1+t^2}\; ,\; dx=\frac{2\, dt}{1+t^2}\; \Big ]=\\\\=\int\limits^{1}_{0}\frac{2\, dt}{(1+\frac{2(1-t^2)}{1+t^2})\cdot (1+t^2)}=\int\limits^1_0\frac{2\, dt}{3-t^2}=\frac{2}{2\sqrt3}\, ln\Big |\frac{\sqrt3+t}{\sqrt3-t}\, \Big |\Big |_0^1=\\\\=\frac{1}{\sqrt3}\cdot \Big (ln\Big |\frac{\sqrt3+1}{\sqrt3-1}\, \Big |-ln|1|\Big )=\frac{1}{\sqrt3}\cdot ln\Big |\frac{\sqrt3+1}{\sqrt3-1}\Big |$