x的平方乘以根号下2x—1分之一的不定积分亲那x的平方是分母

∫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)+2∫1/(t^2+1)*d(1/t)
=-2/t(t^2+1)+2∫1/(t^2+1)*(-1/t^2)*dt
=-2/t(t^2+1)-2∫[1/t^2-1/(t^2+1)]dt
=-2/t(t^2+1)-2∫dt/t^2+2∫dt/(t^2+1)
=-2/t(t^2+1)+2/t+2arctant
=2arctan√(2x-1)+2/√(2x-1)-2/[√(2x-1)*(2x-1+1)]
=2arctan√(2x-1)+2/√(2x-1)-1/[x√(2x-1)]
我的妈呀，这么麻烦

令y² = 2x - 1、y dy = dxx = (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 + y...