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)
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
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)