题目

求不定积分∫ xdx/√(2X^2-4x)

提问时间：2020-11-15

答案

∫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