∫(dx/(x²+x-6))=∫dx/(x²+2*x*(1/2)+1/4-1/4-6)=∫(dx/((x+1/2)²-25/4)=
=∫(dx/(-(5/2)²-(x+1/2)²).
Используем формулу "Высокого логарифма":
∫(dx/(a²-x²)=(1/(2a))*(ln|a+x|/ln|a-x|)+C x≠a
(1/(2*5/2))*(ln|(-5/2+x+1/2)|/ln|(-5/2-x-1/2|)=
=(ln|x-2|/ln|-x-3|)/5==(ln|x-2|/ln|-(x+3)|)/5=(ln|x-2|/ln|x+3|)/5.
Y=x³ y=2+x y=0
x³=0 x=0
2+x=0 x=-2 ⇒
S=int(x³-2-x-0) I₋₂⁰=(x⁴/4-2x-x₂/2) I₋₂⁰=4+4-2=6.
sin2x=2tgx /(1+tg²x)= 2(-4/3) -8/3 -8*9 24
=-------------=-----------= - ----------= - -------
1+(16/9) 25/9 3*25 25
Если П/2<х<П , то П<2х<2П и на этом промежутке синус отрицательный
2x²/(x-1) -3x/(x+2) =2(4x-1)/(x² +x-2) ;* * * ОДЗ : x≠ -2 и x≠1 * * *
(<u>2x² -2 </u>+2)/(x-1) -(<u>3x+6</u> -6)/(x+2) =2(4x-1)/(x² +x-2) ;
2(x² -1)/(x-1) +2/(x-1) -3(x+2)/(x+2) +6/(x+2) =2(4x-1)/(x² +x-2) ;
2(x+1) -3 +2/(x-1) +6/(x+2) =2(4x-1)/(x² +x-2) ;
2(x+1) -3 + <u>(2(x+2)+6(x-1))/(x-1)(x+2)</u> =2(4x-1)/(x² +x-2) ;
2(x+1) -3 + 2(4x-1)/(x² +x-2) =2(4x-1)/(x² +x-2) ;
2(x+1) -3 =0 ;
2x+2 -3 =0 ;
x=1/2.