Question

结果怎么算？要过程！急求！

Answer 1

如果是d²[∫[0：x²][1/(1+t)]dt]/dx²，那么：

∫[0：x²][1/(1+t)]dt
=ln|1+t||0：x²|
=ln|1+x²|-ln|1+0|
=ln(1+x²)-0
=ln(1+x²)
dy/dx=[ln(1+x²)]'=2x/(1+x²)
d²y/dx²=[2x/(1+x²)]'
=[(2x)'(1+x²)-2x·(1+x²)']/(1+x²)²
=[2(1+x²)-2x·2x]/(1+x²)²
=2(1-x²)/(1+x²)²

Answer 2

∫(0,x²)dt/(1+t)
=ln|1+t} (0,x²)
=ln(1+x²)

令y=ln(1+x²)
则dy/dx=1/(1+x²)*2x=2x/(1+x²)
所以原式=d²y/dx²
=[2(1+x²)-2x*2x]/(1+x²)²
=(2-2x²)/(1+x²)²