1个回答
展开全部
分部积分法:∫udv=uv-∫vdu
∫(0,1)ln(1+x²)dx
=xln(1+x²)|(0,1)-∫(0,1)x/(1+x²).2xdx
=(ln2-0)-2∫(0,1)(x²+1-1)/(1+x²).dx
=ln2-2∫(0,1)[1-1/(1+x²)]dx
=ln2-2[x-arctanx](0,1)
=ln2-2[1-π/4-0]
=ln2-2+π/2
=ln2+(π-4)/2
e^[ln2+(π-4)/2]
=e^ln2.e^(π-4)/2
=2e^(π-4)/2
∫(0,1)ln(1+x²)dx
=xln(1+x²)|(0,1)-∫(0,1)x/(1+x²).2xdx
=(ln2-0)-2∫(0,1)(x²+1-1)/(1+x²).dx
=ln2-2∫(0,1)[1-1/(1+x²)]dx
=ln2-2[x-arctanx](0,1)
=ln2-2[1-π/4-0]
=ln2-2+π/2
=ln2+(π-4)/2
e^[ln2+(π-4)/2]
=e^ln2.e^(π-4)/2
=2e^(π-4)/2
本回答被提问者和网友采纳
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