∫d/dx[X^tan(x^2)]dx =∫d[X^tan(x^2)]=X^tan(x^2)+C∫dx/(2+3X^2) =(1/2)*∫dx/(1+(√(3/2)x)^2)=(1/2)*(√(2/3))∫d(√(3/2)x)/(1+(√(3/2)x)^2)=(√6/6)*arctan(√6x/2) +C (∫dx/(1+x^2)=arctanx+C)...第一题的不定积分，两个dx怎么可以约掉呢。。df(x)/dx就是导数了，你的问题我认为是∫d(X^tan(x^2))dx/dx =∫f'(x)dx=f(x)+C