∫dt/(1+t^2)+∫dt/(1+t2 )

题目

∫dt/(1+t^2)+∫dt/(1+t2 )
第一个 x 第二个0~1/x 上下限

答案

∫dt/(1+t^2)+∫dt/(1+t^2)
=arctant|+ractant|
=arctanx+arctan(1/x)
={π/2,x>0;
{-π/2,x