求不定积分∫In(x^2+1)dx

题目

求不定积分∫In(x^2+1)dx

答案

分部积分
∫ln(x²+1)dx
=xln(x²+1)-∫2x²/(x²+1)dx
=xln(x²+1)-2∫[1-1/(x²+1)]dx
=xln(x²+1)-2∫1dx+2∫1/(x²+1)dx
=xln(x²+1)-2x+2arctanx+C