Помогите решить ,интегралы.люди добрые.

Решите задачу:

$1.\,\, \int\limits^{-1}_{-2} {(-\frac{5}{x^2}+x^2-3x)} \, dx =(\frac{5}x+\frac{x^3}3-\frac{3x^2}2)|^{-1}_{-2}=\\=-5-\frac{1}3-\frac{3}2-(-\frac{5}2-\frac{8}{3}-\frac{3*4}{2})=-\frac{41}6+\frac{67}6=\frac{26}6=\frac{13}3$

$2.\,\, \int\limits^8_5 {(\frac{2}{(x-2)^2}-\frac{1}{\sqrt{x-4}}) \, dx =(2*\frac{(x-2)^{-2+1}}{-2+1}-\frac{(x-4)^{-\frac{1}2+1}}{-\frac{1}2+1})|^8_5=$
$=(-\frac{2}{x-2}-2(x-4)^\frac{1}2)|^8_5=-\frac{2}{6}-2*4^{\frac{1}2}-(-\frac{2}{3}-2*1)=\frac{1}{3}-2=-\frac{5}{3}$

$3.\,\, \int\limits^2_0 {\frac{x^3-27}{x^2+3x+9} \, dx =\int\limits^2_0 {\frac{(x-3)(x^2+3x+9)}{x^2+3x+9} \, dx=\int\limits^2_0 {(x-3) \, dx=\frac{(x-3)^2}{2}|^2_0=$
$=\frac{1}2-\frac{9}2=-\frac{8}2=-4$

$4.\,\, \int\limits^2_1 {\frac{x^3-64}{x^2+4x+16} \, dx =\int\limits^2_1 {\frac{(x-4)(x^2+4x+16)}{x^2+4x+16} \, dx=\int\limits^2_1 {(x-4) \, dx=\frac{(x-4)^2}{2}|^2_1=$
$=\frac{4}2-\frac{9}2=-\frac{5}2$

$5.\,\, \int\limits^3_2 {\frac{6x^4-4x^3+7x^2-1}{x^2} \, dx =\int\limits^3_2 {(\frac{6x^4}{x^2}-\frac{4x^3}{x^2}+\frac{7x^2}{x^2}-\frac{1}{x^2}) \, dx=$
$=\int\limits^3_2 {(6x^2-4x+7-\frac{1}{x^2})} \, dx =(2x^3-2x^2+7x+\frac{1}x)|^3_2=\\\\=2*27-18+21+\frac{1}3-(16-8+14+\frac{1}2)=\frac{344}6-\frac{135}6=\frac{209}6$