Question

如图，不定积分运算

Answer 1

设 x = 2tanθ，则 dx = 2sec²θ*dθ, θ = arctan(x/2)。cosθ = 1/√[1+tan²θ] = 4/√(x²+4)，
sinθ = cosθ*tanθ = 4/√(x²+4) * (x/2) = 2x/√(x²+4)
那么，上面的积分变换为：
=∫(2sec²θ * dθ)/[4tan²θ * √(4tan²θ+4)]
=∫2sec²θ*dθ/[4tan²θ*2secθθ]
=1/4*∫secθ*dθ/tan²θ
=1/4*∫secθ*dθ * (cos²θ/sin²θ)
=1/4*∫cosθ*dθ/(sin²θ)
=1/4*∫d(sinθ)/sin²θ
=1/4*(-1/sinθ) + C
=-1/4 * 1/sinθ + C
=-1/4 * √(x²+4) /(2x) + C
=-√(x²+4) /(8x) + C

Answer 2

令x=2tant,换元后可求得

Answer 3

令x=2tant，则dx=2(sect)^2dt
原积分=1/4∫sectdt/(tant)^2
=1/4∫costdt/(sint)^2
=1/4∫d(sint)/(sint)^2
=-sint/4+C
=x/[4√(x^2+4)]+C