Помогите решить интегралы

Решите задачу:

$1) \int\limits^4_1 \frac{x^2-3x+1}{x^2} \, dx = \int\limits^4_1 (1-\frac{3}{x}+x^{-2})\, dx =\\\\=(x-3\cdot ln|x|+\frac{x^{-1}}{-1})\Big |_1^4=(x-3\cdot ln|x|-\frac{1}{x})\Big |_1^4=\\\\=(4-3ln4-\frac{1}{4})-(1-3ln1-1)=3,75-3ln4-0=3,75-6ln2$

$2)\; \; \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} (sinx+4cosx-1) \, dx =(-cosx+4sinx-x)\Big |_{\frac{\pi}{3}}^{\frac{\pi}{2}}=\\\\=(-cos\frac{\pi}{2}+4sin\frac{\pi}{2}-\frac{\pi}{2})-(-cos\frac{\pi}{3}+4sin\frac{\pi}{3}- \frac{\pi}{3} )=\\\\=0+4- \frac{\pi }{2} + \frac{1}{2}-4\cdot \frac{\sqrt3}{2}+ \frac{\pi }{3} =4,5-2\sqrt3-\frac{\pi}{6}$

$3)\; \; \int\limits^{\frac{\pi }{8}}_0 sin(2x+\frac{\pi}{4}) \, dx =- \frac{1}{2} \cdot cos(2x+\frac{\pi }{4})\Big |_0^{\frac{\pi}{8}} =\\\\=- \frac{1}{2} \cdot \Big (cos \frac{\pi}{2} -cos \frac{\pi}{4}\Big )=- \frac{1}{2} \cdot \Big (0-\frac{\sqrt2}{2} \Big ) =\frac{\sqrt2}{4}$