求不定积分:积分号dx⼀根号下(1+x-x^2)

2024-11-30 06:21:30
推荐回答(2个)
回答1:

解:
∫dx/√(1+x-x^2)
=∫dx/[5/4-(x-1/2)^2]
=4/5·∫dx/[1-(√5/2·x-√5/2)^2]
=4/5·2/√5·∫d(√5/2·x)/[1-(√5/2·x-√5/2)^2]
=8√5/25·arcsin(√5/2·x)+C

回答2:

0