Question

求不定积分∫(2x-2)/(x^2+2x+5)dx

Answer 1

∫(2x-2)/(x^2+2x+5)dx
=∫(2x+2-4)/[x²+2x+5]dx
=∫d(x²+2x+5)/(x²+2x+5)-∫4dx/[(x+1)²+4]
=ln(x²+2x+5)-2tan^(-1)[(x+1)/2] +C

Answer 2

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