Question

求不定积分∫ xdx/√(2X^2-4x)

Answer 1

∫ 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

Answer 2

I = ∫xdx/√(2x^2-4x) = (1/√2)∫xdx/√(x^2-2x) = (1/√2)∫xdx/√[(x-1)^2-1]
令 x-1 = sect， 则
I = (1/√2)∫(1+sect) sect dt = (1/√2)∫sect dt + (1/√2)∫(sect)^2 dt
= (1/√2) tant + (1/√2)ln|sect+tant| + C
= √(2x^2-4x) + (1/√2)ln|x-1+√(x^2-2x)| + C