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)
网友回答
∫[(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
======以下答案可供参考======
供参考答案1:
u/v=(u'v-uv')/v^2
??/