Question

数学积分题∫（t+2）/(t^2+2t+2)dt,过程详细

Answer 1

c为任意常数

Answer 2

∫（t+2）/(t^2+2t+2)dt
（t+2）/(t^2+2t+2)
= (t+2)/(t+1)^2
let
(t+2)/(t+1)^2 = A/(t+1) + B/(t+1)^2
=>A(t+1) +B = t+2
=> A = 1 , B =1
∫（t+2）/(t^2+2t+2)dt
= ∫ 1/(t+1) dt + ∫ 1/(t+1)^2 dt
= ln|t+1| - 1/(t+1) + C

追问

= (t+2)/(t+1)^2
不对吧，应该是= (t+2)/{(t+1)^2+1}

Answer 3

解：令x=t+1代入∫（t+2）/(t^2+2t+2)dt得
∫（t+2）/(t^2+2t+2)dt= ∫（x+1）/(x^2+1)dx
=∫x/(x^2+1)dx+∫1/(x^2+1)dx
=1/2∫1/(x^2+1)d(x^2+1)+∫1/(x^2+1)dx
=1/2ln(x^2+1)+arctan(x)+C
=1/2ln(t^2+2t+2)+arctan(t+1)+C