Question

x∧2/(1+e∧x)不定积分可以求吗

Answer 1

不定积分不行,初等函数不能表达.如果是定积分.对称区间积分,还能做.
举例：
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] x² dx
=(1/6)x³ |[-1-->1]
=(1/6)-(-1/6)
=1/3