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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,
人气:214 ℃ 时间:2020-05-23 06:29:07
解答
∫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
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