因为：(arctanx)'=1/(1+x^2)故(arctankx)'=1/(1+k^2x^2)*k=k/(1+k^2x^2) 分子分母都除以k^2得=1/k*1/(x^2+1/k^2)于是(karctankx)'=1/(x^2+1/k^2)令1/k^2=2解得k=√2/2故∫1/（x^2+2）dx=√2/2*arctan(√2x/2)+C
