Question

积分 dx/√(x^2+2x+2)用欧拉代换怎么求

Answer 1

√(x^2+2x+2)+t=x

(x-t)^2=x^2+2x+2

t^2-2=(2+2t)x

x=(t^2-2)/(2+2t)

dx=(2t*(2+2t)-(t^2-2)*2)/(2+2t)^2dt=(2t^2+4t+4)/(2+2t)^2dt

原式=∫1/((t^2-2)/(2+2t)-t)*(2t^2+4t+4)/(2+2t)^2dt

=∫1/((t^2-2)/(2+2t)-t)*(2t^2+4t+4)/(2+2t)^2dt=∫-dt/(1+t)

=-ln|1+t|+C

=-ln|1+x-√(x^2+2x+2)|+C