求不定积分∫xln(1+x^2)dx
展开全部
∫xln(1+x^2)dx
=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
=1/2∫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
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
