Question

高等数学，不定积分问题，求解题思路与步骤

Answer 1

原式=1/2*∫2(x+1-2)dx/(x²+2x+3)
=1/2*∫(2x+2)dx/(x²+2x+3)-1/2*∫4dx/(x²+2x+3)
=1/2*∫d(x²+2x+3)/(x²+2x+3)-2∫d(x+1)/[(x+1)²+2]
=1/2*ln|x²+2x+3|-√2*arctan[(x+1)/√2]+C

Answer 2

[ln(x^2+2x+3) ]'= (2 x + 2) / (x^2 + 2 x + 3)

[arctan(x)]' = 1/ ( 1 + x^2)
[arctan( (x+1) / a )]' = 1/ [ a *( 1 + ((x+1)/a)^2))]
积分 =0.5* ln(x^2+2x+3) - a * arctan( (x+1) / a ) +C
a = Sqrt(2)