∫ln(1+x^2)dx=xln(1+x^2)-∫ 2x^2 / (1+x^2) dx
又 ∫ 2x^2 / (1+x^2) dx=2∫ [1-1/(1+x^2)]dx=2x-2acrtanx
meili7085a 精英
共回答了369个问题采纳率:88.3% 举报
=1/2∫ln(1+x^2)dx^2
=1/2∫ln(1+x^2)d(1+x^2)
=1/2(1+x^2)ln(1+x^2)-1/2∫(1+x^2)dln(1+x^2)
=1/2(1+x^2)ln(1+x^2)-1/2∫(1+x^2)*1/(1+x^2)d(1+x^2)
=1/2(1+x^2)ln(1+x^2)-1/2∫dx^2
=1/2(1+x^2)ln(1+x^2)-1/2x^2+C
1年前
你能帮帮他们吗