Question

求不定积分∫1/^3+2x-x²*dx

Answer 1

∫[1/(3+2x-x²)]dx = ∫dx/[(1+x)(3-x)] = (1/4)∫[1/(1+x) + 1/(3-x)]dx
= (1/4)[ln|1+x| - ln|3-x|] + C = (1/4)ln|(1+x)/(3-x)| + C

Answer 2

∫1/(3+2x-x²)]dx

= ∫dx/[(1+x)(3-x)]

=¼∫[1/(1+x) + 1/(3-x)]dx

=¼[ln|1+x| - ln|3-x|] + C

=¼ln|(1+x)/(3-x)| + C