1/根号下(1+x^2),求积分
1个回答

令x=tana

a=arctanx

seca=√(x²+1)

1+x²=sec²a

dx=sec²ada

原式=∫sec²ada/seca

=∫secada

=∫(1/cosa)da

=∫[cosa/cos²a]da

=∫d(sina)/(1-sin²a)

=(1/2)∫[1/(1-sina)+1/(1+sina)]d(sina)

=(1/2)[-ln|1-sina|+ln|1+sina|]+C

=(1/2)ln|(1+sina)/(1-sina)|+C

=ln|seca-tana|+C

=ln|√(x²+1)-x|+C