一道积分问题...跪求解题过程...万分感谢

∫(x-1)dx/[x²√(2x²-2x+1)]
= (1/√2) * ∫(x-1)dx/[x²√(x² - x + 1/2)]
令x = (tant + 1)/2，则dx = (sec²t)/2 dt
∫(x-1)dx/[x²√(x² - x + 1/2)]
= ∫(tant - 1)/2 * (sec²t)/2 dt * 8cost/(tant + 1)²
= 2∫(tant - 1)dt/[cost(tant + 1)²]
= 2∫(sint - cost)dt/(sint+cost)²
而(sint - cost)dt = -d(sint + cost)
则∫(x-1)dx/[x²√(x² - x + 1/2)]
= -2∫d(sint + cost)/(sint + cost)²
= 2/(sint + cost) + C

∫(x-1)dx/[x²√(2x²-2x+1)]
= √2/(sint + cost) + C
其中t = arctan(2x-1)
整理得∫(x-1)dx/[x²√(2x²-2x+1)]
= [√(2x²-2x+1)]/x + C