用换元法求不定积分的题 ∫x⼀√(5+x-x^2),

2024-12-05 07:44:54
推荐回答(2个)
回答1:

简单分析一下即可,详情如图所示

回答2:

x=-1/2×(1-2x)+1/2,所以
∫xdx/√(5+x-x^2)
=-1/2×∫(1-2x)dx/√(5+x-x^2)+1/2×∫dx/√(5+x-x^2)
=-1/2×∫1/√(5+x-x^2) d(5+x-x^2) + 1/2×∫1/√[21/4-(x-1/2)^2]
=-√(5+x-x^2)+1/2×arcsin[(2x-1)/√(21)]+C