Задача 25979 4x+1/(3x^2+7x-1)^1/2dx...
Условие
предмет не задан
371
Решение
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x+(7/6)=u ⇒ 4x+1=4*(u-(7/6))+1=4u-(22/6)=4u-(11/3)
dx=du
∫ (4x+1)dx/sqrt(3x^2+7x–1)=
=(1/sqrt(3))* ∫ (4u-(11/3))du/sqrt(u^2-(61/36))=
=(1/sqrt(3))* ∫ 4udu/sqrt(u^2-(61/36))+
+(11sqrt(3)/9)* ∫ du/sqrt(u^2-(61/36))=
=(2/sqrt(3))*2sqrt(u^2-(61/36)) +
+(11sqrt(3)/9)*ln|u+sqrt(u^2-(61/36))|+C=