定积分(x+1)dx/(x^2+2x+2),被积函数属于-2到0,_数学解答

定积分(x+1)dx/(x^2+2x+2),被积函数属于-2到0,

人气：410 ℃　时间：2019-10-30 14:26:28

解答

∫(- 2→0) (x + 1)/(x² + 2x + 2) dx
= (1/2)∫(- 2→0) (2x + 2)/(x² + 2x + 2) dx
= (1/2)∫(- 2→0) d(x² + 2x + 2)/(x² + 2x + 2)
= (1/2)ln(x² + 2x + 2) |(- 2→0)
= (1/2)ln(2) - (1/2)ln(4 - 4 + 2)
= (1/2)ln(2) - (1/2)ln(2)
= 0