1 зд и 2 зд
x^2+7x+11=x^2+2*x*(7/2)+(49/4)-(49/4)+11=(x+(7/2))^2-(5/4)
Табличный интеграл
∫ du/(u^2-a^2)=(1/(2a)) * ln|(x-a)/(x+a)|+C
u=(x+(7/2)
du=dx
a^2=5/4
a=sqrt(5)/2
Получим тот ответ, который и написан
2.
x^3=x*x^2
x^3*sqrt(1-x^2)=x*x^2*sqrt(1-x^2)=-x*(-x^2)*sqrt(1-x^2)=
=-x*(1-x^2-1)sqrt(1-x^2)= -x*(1-x^2)*sqrt(1-x^2)+x*sqrt(1-x^2)
∫x^3*sqrt(1-x^2)dx= ∫ [b]([/b]-x*(1-x^2)*sqrt(1-x^2)+x*sqrt(1-x^2) [b])[/b]dx=
= (1/2)∫ (-2x)*(1-x^2)^(3/2) -(1/2)* ∫ (-2x)(1-x^2)^(1/2)dx=
=(1/2) ∫ (1-x^2)^(3/2)d(1-x^2) -(1/2)* ∫ (1-x^2)^(1/2)d(1-2x^2)=
=(1/2) ∫ (u^(3/2))du -(1/2) ∫ u^(1/2)du=
=(1/2) *(1-x^2)^(5/2)/(5/2) - (1/2) * (1-x^2)^(3/2)/(3/2)+C=
=(1/5)sqrt((1-x^2)^5)-(1/3)sqrt((1-x^2)^3)+C
в ответе не должно быть х после (1/3)