题目

积分(1-cosx)dx/(x-sinx)
积分[(1-cosx)dx]/(x-sinx)
要具体的啊..
上面那题 不要做了.要做也可以 .
换成这题
积分[(x-3)dx]/[(x^2)+2x+3)

提问时间：2020-07-24

答案

∫[(1-cosx)dx]/(x-sinx)
=∫d(x-sinx)/(x-sinx)
=ln(x-sinx)+C
原式=∫(x+1-4)dx/(x²+2x+3)
=∫(x+1)dx/(x²+2x+3)-∫4dx/(x²+2x+3)
=1/2∫(2x+2)dx/(x²+2x+3)-∫4dx/[(x+1)²+2]
=1/2∫d(x²+2x+3)dx/(x²+2x+3)-2∫dx/[(x+1)²/2+1]
=1/2*ln(x²+2x+3)-2∫dx/[(x/√2+1/√2)²+1]
=1/2*ln(x²+2x+3)-2√2∫d(x/√2+1/√2)/[(x/√2+1/√2)²+1]
=1/2*ln(x²+2x+3)-2√2arctan(x/√2+1/√2)+C