Решение
(tg4x - tg3x) / (1 + tg4x*tg3x) = √3
tg((4x - 3x) = √3
tgx = √3
x = arctg√3 + πn, n ∈ Z
x = π/3 + πn, n ∈ Z
Применили формулу: tg(α - β) = (tgα - tgβ) / ( 1 + tgα * tgβ)
∫(5+х)/(3x^2+1)dx
∫(x/3x²+1)+(56/3x²+1)dx={u=3x²+1; du=6xdx;dx=du/6x}=1/6∫du/u+5∫dx/(3x²+1)=
=logu/6+{s=√3dx}=logu/6+5/√3∫ds/(s²+1)=5tg⁻¹(s)/√3+logu/6=
=1/6log(3x²+1)+5tg⁻¹(√3x)/√3+c
<span>(с-2)(с-3)-(с+1)(с+3)= (c^2+3c-2c+6-c^2+3c-c+3)/(c-3)(c+3)=(3c+6)/(c^2-3^2=3(c+2)/c^2-3^2</span>