用换元积分法求∫dx/(x^2+6x+10)

题目

用换元积分法求∫dx/(x^2+6x+10)
我知道答案是arctan（x+3）,所以需要具体的解答过程,

答案

∫dx/(x^2+6x+10)=∫dx/[(x+3)^2+1]
=∫d(x+3)/[(x+3)^2+1] 令t=x+3
=∫dt/(t^2+1)
=arctant+C
=arctan（x+3)+C