求不定积分∫ xdx/√(2X^2-4x)

∫x/√(2x²-4x) dx
= ∫x/√[(√2*x-√2)²-2] dx
令√2*x-√2=√2*secy,x-1=secy,dx=secy*tany dy
cosy=1/(x-1),siny=√[(x-1)²-1] / (x-1) = √(x²-2x) / (x-1)
原式= ∫(1+secy)/(√2*tany) * (secy*tany) dy
= (1/√2)∫(1+secy)(secy) dy
= (1/√2)∫(sec²y+secy) dy
= (1/√2)tany + (1/√2)ln|secy+tany| + C
= (1/√2)√(x²-2x) + (1/√2)ln|x-1+√(x²-2x)| + C

∫ xdx/√(2X^2-4x)
=√2/2∫ xdx/√(X^2-2x)
=√2/2∫ (x-2+2)dx/√(X^2-2x)
=√2/2∫ [(x-1)/√(X^2-2x)+1/√(X^2-2x)]dx
=√2/4∫ [1/√(X^2-2x)d(x^2-2x)+√2/2∫ 1/√(X^2-2x)]dx
=√2/2*√(X^2-2x)++√2/2∫ 1/√(X^2-2x)]dx
剩下的一项套公式吧
∫1/√(x^2-a^2)dx