Пошаговое объяснение:
a) F(x) = ∫ ( (3x-2)²-(1-2x)³ ) dx == ∫ ((3x)² - 2*3x*2 + 2² - 1(1³ -3*1²*2x + 3*1*(2x)² - (2x)³)) dx =
= ∫ (9x² - 12x + 4 - (1 - 6x + 12x² - 8x³)) dx =
= ∫ (9x² -12x +4 - 1 +6x -12x² +8x³) dx =
= ∫ (8x³ - 3x² -6x +3) dx =
= ∫ (8x³) dx + ∫ (-3x²) dx + ∫ -6x dx + ∫ 3dx =
= 8 ∫x³ dx - 3 ∫x² dx -6 ∫x dx +3 ∫dx =
= 8*x⁴/4 - 3*x³/3 - 6*x²/2 +3x/1 + C =
= 2x⁴ - x³ - 3x² +3x + C, C=const
b) F(x) = ∫ (√(3x+1)) dx == [пусть 3х+1=а, тогда da=(a)' * dx =
= (3x+1)' * dx = 3dx → dx = da / 3] =
= ∫ 1/3 √a da = (1/3) * ∫ √a da =
= (1/3) * ∫ a^(1/2) da =
= (1/3)*(a^(1/2 + 1) / (1/2+1) ) + C=
=(1/3) * (a * √a) / (3/2) + C =
= (1/3) * (a√a) * (2/3) + C =
= 2a√a / 9 + C =
= [ a=3x+1] =
= 2*(3x+1)√(3x+1) / 9 + C, C=const
