求∫1/[x^2根号（x^2+1）]dx的不定积分

x=tanadx=sec²ada原式=∫sec²ada/tan²a*seca=∫secada/tan²a=∫da/sina∫sinada/sin²a=∫dcosa/(cos²a-1)=(1/2)[∫dcosa/(cosa-1)-∫dcosa/(cosa+1)]=(1/2)(ln|cosa-1|-ln|cosa+1|)+C...