Question

不定积分∫ln(1+x^2)dx

Answer 1

用分部积分法,

(uv)'=u'v+uv',

设u=ln(1+x^2),v'=1,

u'=2x/(1+x^2),v=x,

原式=xln(1+x^2)-2∫x^2dx/(1+x^2)

=xln(1+x^2)-2∫(1+x^2-1)dx(1+x^2)

=xln(1+x^2)-2∫dx+2∫dx/(1+x^2)

=xln(1+x^2)-2x+2arctanx+C.