∫[0,1]1/(x^2+6x+9)dx
=-1/(x+3)[0,1]
=1/12这个-1/(x+3）是怎样得出来的∫[0,1]1/(x^2+6x+9)dx=∫[0,1]1/(x+3)^2dx=∫[0,1]1/(x+3)^2d(x+3)=-1/(x+3)[0,1]=1/12