∫(2√x⁴-3x²)dx/x²=∫(x^(3/4-2)dx-3∫dx=∫dx/x^(1,25)-3x+C=∫x^(-5/4)dx-3x+C=
-4·x^(-1/4)-3x+C
-4·1/√1-3·1+C=4, -4-3+C=4, C=11
Ответ:-4/⁴√x-3x+11
∫(x²dx/(5x³+2)
Пусть 5х³+2=t,тогда dt=15x²dx, dx=dt/15x²=dt/15t
∫(x²dx/(5x³+2)=∫dt/15t=15Lnt+C= 15Ln(5x³+2)+C
π/3 π/3
∫cosxdx=-4sinxl = -4(sinπ/3-sinπ/2)=-4(√3/2+1)=4-2√3
π/2 π/2