√(1+x^2 )的 不定积分怎么求?（根号下1加上x的平方）

Answer 1

∫√(1+x^2 )dx
令x=tant，
原式=∫sect·dtant (注：本式还等于∫sec³tdt)

=sect·tant-∫tantdsect
=sect·tant-∫tant·tantsectdt
=sect·tant-∫(sec²t-1)sectdt
=sect·tant-∫(sec³t-sect)dt
=sect·tant-∫sec³tdt+∫sectdt
=sect·tant-∫sect·dtant +∫sectdt
所以
2×∫sect·dtant=sect·tant-∫sect·dt
=sect·tant-ln|sect+tant|+2c
=x√(1+x²)-ln|x+√(1+x²)|+2c
即
原式=1/2x√(1+x²)-1/2ln|x+√(1+x²)|+c

Answer 2

x = sinθ,dx = cosθ dθ ∫ √(1 - x²) dx = ∫ √(1 - sin²θ)(cosθ dθ) = ∫ cos²θ dθ = ∫ (1 + cos2θ)/2 dθ = θ/2 + (sin2θ)/4 + C = (arcsinx)/2 + (sinθcosθ)/2 + C = (arcsin

Answer 3

Answer 4

"所以"那一步后面，两个三角函数之间应该是加号不是减号

Answer 5

原式=∫sect·dtant (注：本式还等于∫sec³tdt)
=sect·tant-∫tantdsect
=sect·tant-∫tant·tantsectdt
=sect·tant-∫(sec²t-1)sectdt
=sect·tant-∫(sec³t-sect)dt
=sect·tant-∫sec³tdt+∫sectdt
=sect·tant-∫sect·dtant +∫sectdt
所以
2×∫sect·dtant=sect·tant＋∫sect·dt
=sect·tant＋ln|sect+tant|+2c
=x√(1+x²)＋ln|x+√(1+x²)|+2c
即
原式=1/2x√(1+x²)＋1/2ln|x+√(1+x²)|+c