Question

∫1/x²-x-6dx

不定积分

Answer 1

原式=∫1/xln²xdx+∫x²ln²xdx=∫ln²xdlnx+∫ln²xdx³/3=1/3ln³x+ln²x·x³/3-1/3∫x³·2lnx·1/xdx=1/3ln³x+ln²x·x³/3-2/3∫x²lnxdx=1/3ln³x+ln²x·x³/3-2/3∫lnxdx³/3=1/3ln³x+ln²x·x³/3-2/9x³lnx+2/9∫x²dx=1/3ln³x+ln²x·x³/3-2/9x³lnx+2/27x³+c

Answer 2

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