第4个回答 2019-09-22
这题对1/(u^2+a^2)反分布积分得到的
因为∫1/(u^2 +a^2)du = u/(u^2+a^2) - ∫ud(1/(u^2+a^2))
=u/(u^2+a^2) +∫2u^2/(u^2+a^2)^2 du
=u/(u^2+a^2) +2∫ (u^2 +a^2-a^2)/(u^2+a^2)^2 du
=u/(u^2+a^2)+2∫1/(u^2+a^2) -a^2/(u^2+a^2)^2 du
既A=∫1/(u^2 +a^2)du =u/(u^2+a^2)+2∫1/(u^2+a^2) -a^2/(u^2+a^2)^2 du
=u/(u^2+a^2)+2A-2∫a^2/(u^2+a^2)^2 du
A=∫1/(u^2 +a^2)du=-u/(u^2+a^2) +2∫a^2/(u^2+a^2)^2 du
∫a^2/(u^2+a^2)^2 du = 1/(2a^2) * [A+u/(u^2+a^2)]