Помогите пожалуйста, прошу, срочно )

Решите задачу:

$\displaystyle 1. \int^3_2 \frac{1+x^5}{x^4}dx= \int^3_2 \frac{x^5}{x^4}dx+ \int^3_2 \frac{1}{x^4}dx= \int^3_2xdx- \frac{1}{3x^3}=$
$\displaystyle = \frac{x^2}{2}- \frac{1}{3x^3}= \frac{3x^5-2}{6x^3} \bigg|^3_2= \frac{729-2}{162}- \frac{96-2}{48} = \frac{727}{162}- \frac{94}{48}=$
$\displaystyle = \frac{34896-15228}{7776}= \frac{6556}{2592}= \frac{1639}{648}$

$\displaystyle 2. \int^{ \frac{\pi}{6}}_{ \frac{\pi}{8}} 2 \cdot \frac{1}{cos^2x}dx=2 \cdot \frac{1}{2}tg2x=tg2x \bigg|^{ \frac{\pi}{6}}_{ \frac{\pi}{8}} =tg \frac{\pi}{3}-tg \frac{\pi}{4}= \sqrt{3}-1$

$\displaystyle 3. \int^1_0 x^2(2-x^3)^4dx= \frac{(x^3-2)^5}{15} \bigg|^1_0=- \frac{1}{15}-(- \frac{32}{15})= \frac{31}{15}$