求∫(arctanx/x^2)dx,下面是我算的,答案怎么是(-1/x)*arctanx-(1/2)ln(1+1/x^2)+C设u=arctanx,1/x^2dx=d(1/-x)=dv∫(arctanx/x^2)dx=(1/-x)arctanx-∫(1/-x)*d(arctanx)=(1/-x)arctanx+∫1/(1+x^2)*d(lnlxl)=(1/-x)arctanx+∫(1+x^2-x^2)/(1+x^2)*d

Question

求∫(arctanx/x^2)dx,下面是我算的,答案怎么是(-1/x)*arctanx-(1/2)ln(1+1/x^2)+C
设u=arctanx,1/x^2 dx=d(1/-x)=dv
∫(arctanx/x^2)dx
=(1/-x)arctanx-∫(1/-x)*d(arctanx)
=(1/-x)arctanx+∫1/(1+x^2)*d(ln lxl)
=(1/-x)arctanx+∫(1+x^2-x^2)/(1+x^2)*d(ln lxl)
=(1/-x)arctanx+ln lxl-(1/2)ln l1+x^2l+C

Answer 1

你和答案是一样的啊
ln|x| - (1/2)ln(1+x^2)
= 1/2[lnx^2 - ln(1+x^2)]
=1/2ln[x^2 / (1+x^2)]
=-1/2 ln[(1+x^2)/x^2]
=-1/2 ln(1+1/x^2)