急,求不定积分∫In(x^2+1)dx ∫下面是0 上面是3
问题描述：
急,求不定积分∫In(x^2+1)dx ∫下面是0 上面是3
答
∫(0->3) In(x^2+1)dx = [xln(x^2+1)](0->3) - ∫(0->3) 2x^2/(x^2+1) dx=3ln10 -∫(0->3) [2 - 2/(x^2+1)] dx=3ln10 - 2[ x- arctanx] (0->3)=3ln10 - 2(3 - arctan3)