Question

部分分式的积分问题

怎么求都求不出来啊，用代换u=x+2也不行

Answer 1

你要给我点悬赏分！那么辛苦啊.
先计算∫dx/(x^2+4x+8)
=∫dx/[(x+2)^2+4]
=∫dx/4[(x/2+1)^2+1]
=(1/2)∫d(x/2+1)/[(x/2+1)^2+1]
=(1/2)arctan(x/2+1)+C
故∫（x+1）dx/(x^2+4x+8)
=∫[（2x+4）/2-1]dx/(x^2+4x+8)
=∫（2x+4）/2dx/(x^2+4x+8)-∫dx/(x^2+4x+8)
=∫1/2d(x^2+4x+8)/(x^2+4x+8)-∫dx/(x^2+4x+8)
=（1/2）ln(x^2+4x+8)-(1/2)arctan(x/2+1)+C

Answer 2

解：
∫(x+1)dx/(x²+4x+8)
=(1/2)∫(2x+4)dx/(x²+4x+8)-∫dx/(x²+4x+8)
=(1/2)∫(x²+4x+8)'dx/(x²+4x+8)-∫dx/(x²+4x+8)
=(1/2)ln(x²+4x+8)-∫d(x+2)/((x+2)²+2²)
=(1/2)ln(x²+4x+8)-(1/2)arctan((x+2)/2)+c