求∫1/[x^2根号（x^2+1）]dx的不定积分

x=tana
dx=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
=(1/2)ln|(cosa-1)/(cosa+1)|+C
=(1/2)ln|(cosa-1)²/sin²a|+C
=ln|(cosa-1)/sina|+C
=ln|cota-csca|+C
=ln|1/x-√[1+(1/x)²]|+C
=ln|[1-√(x²+1)]/x|+C