x的平方乘以根号下2x—1分之一的不定积分

令y² = 2x - 1、y dy = dx
x = (1 + y²)/2、x² = (1 + y²)²/4、1/x² = 4/(1 + y²)²
∫ 1/[x²√(2x - 1)] dx
= ∫ 4/(1 + y²)² * 1/y * y dy
= 4∫ dy/(1 + y²)²、y = tanθ、dy = sec²θ dθ
= 4∫ sec²θ/sec⁴θ dθ
= 4∫ cos²θ dθ
= 2∫ (1 + cos2θ) dθ
= 2θ + sin2θ + C
= 2θ + 2sinθcosθ + C
= 2arctan(y) + 2y/(1 + y²) + C
= 2arctan√(2x - 1) + √(2x - 1)/x + C

∫dx/x^2√(2x-1)
令√(2x-1)=t
x=(t^2+1)/2
dx=2tdt/2=tdt
所以原式=∫tdt/[(t^2+1)/2]^2t
=4∫dt/(t^2+1)^2
=4∫tdt/(t^2+1)^2t
=2∫d(t^2+1)/(t^2+1)^2t
=-2∫d(1/(t^2+1) /t
=-2/t(t^2+1...