补考的高数题~2点半考了~1.已知:f(x)=x三次方+ax平方+bx在x=1处有极值-2,求常数a,b及f(x).4

f(x)=x³+ax²+bx
f'(x)=3x²+2ax+b
f(1)=1+a+b=-2
f'(1)=3+2a+b=0
得a=0、b=-3
f(x)=x³-3x
令√t=u,t=u²,dt=2udu
∫1/t(1+√t)dt
=∫2u/u²(1+u)dt
=2∫1/(u²+u)dt
=2∫1/[(u+1/2)²-(1/2)²]dt
=2ln|(x-1/2)/(x+1/2)||
=2[ln(7/9)-ln(1/3)]
=ln49-ln81+ln9
∫e^|x|dx
=∫e^xdx+∫e^(-x)dx
=∫e^xdx-∫e^(-x)d(-x)
=e^x|-e^(-x)|
=e³-1-1+e²
=e³+e²-2