Question

如何算瑕积分

Answer 1

原式
=∫dx/√(x²+x)
=∫dx/√(x²+x+1/4-1/4)
=∫dx/√[(x+1/2)²-1/4]
=2∫dx/√[4(x+1/肆凯2)²-1]
=2∫dx/√[(2x+1)²-1]
令2x+1=secu
当x=0时敬前，u=0
当x=1时，u=arccos(1/3)
x=(secu-1)/2
dx=secutanudu/2
sec²u-1=tan²u
原式
=∫secutanudu/tanu
=∫tanudu
=∫sinudu/cosu
=-∫d(cosu)/裂稿唤cosu
=-ln|cosu|+C |(arccos(1/3)，0)
=-ln(1/3)+0
=ln3

追问

那是瑕积分，应该要考虑极限吧