Пошаговое объяснение:
∫( f(x)=
cos
2
(3x+1)
2
−3sin(4−x)+2x
F(x) = \int{(\frac{2}{cos^2(3x+1)}-3sin(4-x)+2x)}\, dx=F(x)=
cos
2
(3x+1)
2
−3sin(4−x)+2x)dx= 2\int{\frac{1}{cos^2(3x+1)}\, dx-3\int{sin(4-x)}\, dx+2\int{x}\, dx= \frac{2}{3}\int{\frac{1}{cos^2(3x+1)}\, d(3x+1)+3\int{sin(4-x)}\, d(4-x)+2\int{x}\, dx= \frac{2}{3}tg(3x+1)-3cos(4-x)+x^2
