令u=x+1
则du=dx
原式=∫1/(2+u²2)du
=1/√2·arctan(u/√2)+C
=1/√2·arctan[(x+1)/√2]+C令u=x+1
则du=dx
原式=∫1/(2+u^2)du
=1/√2·arctan(u/√2)+C
=1/√2·arctan[(x+1)/√2]+C∫1/(a^2+x^2)dx
=1/a·arctan(x/a)+C