∫x[ln(x²+2)-ln(2x+1)]dx

∫x[ln(x²+2)-ln(2x+1)]dx
=∫xln(x²+2)dx-∫xln(2x+1)dx
=(1/2)∫ ln(x²+2)d(x²)-(1/2)∫ ln(2x+1)d(x²)
=(1/2)x²ln(x²+2)-(1/2)∫ x²/(x²+2)d(x²)-(1/2)x²ln(2x+1)+(1/2)∫ 2x²/(2x+1)dx
=(1/2)x²ln(x²+2)-(1/2)∫ (x²+2-2)/(x²+2)d(x²)-(1/2)x²ln(2x+1)+(1/4)∫ (4x²-1+1)/(2x+1)dx
=(1/2)x²ln(x²+2)-(1/2)∫1dx+∫ 1/(x²+2)d(x²)-(1/2)x²ln(2x+1)+(1/4)∫(2x-1)dx+(1/4)∫1/(2x+1)dx
=(1/2)x²ln(x²+2)-x/2+ln(x²+2)-(1/2)x²ln(2x+1)+(1/4)(x²-x)+(1/8)ln(2x+1)+C
=(1/2)x²ln(x²+2)+ln(x²+2)-(1/2)x²ln(2x+1)+(1/4)x²-(3/4)x+(1/8)ln(2x+1)+C