Question

(x-1)➗(x^2+4x+5)的不定积分

Answer 1

x^2+4x+5= (x+2)^2 +1

let

x+2 = tanu

dx=(secu)^2 du

∫ (x-1)/(x^2+4x+5) dx 

=(1/2)∫ (2x+4)/(x^2+4x+5) dx  - ∫ 3/(x^2+4x+5) dx

=(1/2)ln|x^2+4x+5|  - 3∫ (secu)^2 du/ (secu)^2

=(1/2)ln|x^2+4x+5|  - 3∫ du

=(1/2)ln|x^2+4x+5|  - 3u + C

=(1/2)(1/2)ln|x^2+4x+5|  - 3arctan(x+2) + C

=(1/2)ln|x^2+4x+5| -∫ 3/(x^2+4x+5) dx

Answer 2

∫(x-1)dx/(x^2+4x+5)
= (1/2)∫(2x+4-6)dx/(x^2+4x+5)
= (1/2)∫d(x^2+4x+5)/(x^2+4x+5) - 3∫d(x+2)/[(x+2)^2+1]
= (1/2)ln(x^2+4x+5) - 3arctan(x+2) + C