Question

Найти неопределенный интеграл (2x-13)/(корень3x^2 - 3x-16)dx

Answer 1

$int frac{2x-13}{sqrt{3x^2-3x-16}}dx=[; 3x^2-3x-16=3(x^2-x)-16=\\=3(x-frac{1}{2})^2-frac{3}{4}-16=3(x-frac{1}{2})^2-frac{67}{4}=3cdot Big ((x-frac{1}{2})^2-frac{67}{12}Big ) ; ]=\\=frac{1}{sqrt3}cdot int frac{2x-13}{sqrt{(x-frac{1}{2})^2-frac{67}{12}}}dx=[; x- frac{1}{2}=t; ,; x=t+frac{1}{2}; ,dx=dt; ]=\\= frac{1}{sqrt3} cdot int frac{2t-12}{sqrt{t^2- frac{67}{12} }} dt=frac{1}{sqrt3}cdot int frac{d(t^2-frac{67}{12})}{sqrt{t^2- frac{67}{12} }}dt -frac{1}{sqrt3}cdot int frac{dt}{sqrt{t^2-frac{67}{12}}}dt=$

$= frac{1}{sqrt3}cdot 2sqrt{t^2- frac{67}{12} } - frac{1}{sqrt3} cdot lnBig |t+sqrt{t^2-frac{67}{12}}Big |+C=\\=frac{2}{sqrt3}cdot sqrt{(x-frac{1}{2})^2-frac{67}{12} }- frac{1}{sqrt3}cdot lnBig |x-frac{1}{2}+sqrt{(x-frac{1}{2})^2-frac{67}{12}}Big |+C$