Question

帮我看看高数不定积分这道题怎么求的。急急急，谢谢大神们

帮我看看高数不定积分这道题怎么求的。急急急，谢谢大神们过程详细一点

Answer 1

令 x = sinu，则
∫dx/[1+√(1-x^2)] = ∫cosudu/(1+cosu) = ∫[1-1/(1+cosu)]du
= u - ∫du/[2cos^2 (u/2)] = u - ∫[sec(u/2)]^2 d(u/2)
= u - tan(u/2) + C = u - (1-cosu)/sinu + C
= arcsinx - [1-√(1-x^2)]/x + C

Answer 2

∫ 1 / [ 1 + √(1-x^2) ] dx
= ∫ 1 / [ 1 + √(1-x^2) ] * [ 1 - √(1-x^2) ] / [ 1 - √(1-x^2) ] dx
= ∫ [ 1 - √(1-x^2) ] / x^2 dx
= ∫ [ 1/x^2 - √(1-x^2) / x^2 ] dx
= - 1/x + √(1 - x^2) / x + arcsinx + C

Answer 3

看不清…………