Question

（x+1）/（x²-2x+5）的不定积分怎么求

Answer 1

原式=∫(x-1)/(x2-2x+5)dx+∫2/(x2-2x+5)dx
=1/2ln(x2-2x+5)+arctan((x-1)/2)+C

Answer 2

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