①∫x^2ln2xdx
=1/3ln2xx^3-1/3∫x^3/2x*2dx
=1/3ln2xx^3-1/9x^3+C
②y=∫[0,3x] t√(1+t^2)dt
=1/2∫[0,3x] √(1+t^2)dt^2
=1/2*2/3*(1+t^2)^(3/2)[0,3x]
=1/3*(1+9x^2)^(3/2)-1/3
y=2x^3+14x-7
y'=6x^2+14=0
x无实根,
因此极值点不存在,在(-∞,+∞)都属于单增
①∫x^2ln2xdx
=1/3ln2xx^3-1/3∫x^3/2x*2dx
=1/3ln2xx^3-1/9x^3+C
②y=∫[0,3x] t√(1+t^2)dt
=1/2∫[0,3x] √(1+t^2)dt^2
=1/2*2/3*(1+t^2)^(3/2)[0,3x]
=1/3*(1+9x^2)^(3/2)-1/3
y=2x^3+14x-7
y'=6x^2+14=0
x无实根,
因此极值点不存在,在(-∞,+∞)都属于单增