∫1/(9+X ²）dX设X=3tant则dX=3sce ²tdtt=arctan（x/3）∫1/(9+X ²）dX=∫1/(9+9tan²t）*3sce ²tdt=∫1/9sce ²t*3sce ²tdt=∫1/3dt=1/3t+C=1/3arctan（x/3）+C