Question

3分之x³的导数为什么是x²

Answer 1

如图所示

Answer 2

令x=-u，则dx=-du，u：1--->-1∫[-1-->1]x2/(1+e^x)dx=-∫[1-->-1]u2/(1+e^(-u))du=∫[-1-->1]u2e^u/(e^u+1)du积分变量可以随便换字母=∫[-1-->1]x2e^x/(e^x+1)dx则得到：∫[-1-->1]x2/(1+e^x)dx=∫[-1-->1]x2e^x/(e^x+1)dx因此∫[-1-->1]x2/(1+e^x)dx=1/2[∫[-1-->1]x2/(1+e^x)dx+∫[-1-->1]x2e^x/(e^x+1)dx]=1/2∫[-1-->1][x2/(1+e^x)+x2e^x/(e^x+1)]dx=1/2∫[-1-->1]x2dx=(1/6)x3|[-1-->1]=(1/6)-(-1/6)=1/3