∫x²ln(1+x²)dx
=∫ln(1+x²)d(x³/3)
=(1/3)x³ln(1+x²)-(1/3)∫x³d[ln(1+x²)]
=(1/3)x³ln(1+x²)-(2/3)∫[x^4/(1+x²)]dx
=(1/3)x³ln(1+x²)-(2/3)∫[x²+1/(1+x²)-1]dx
=(1/3)x³ln(1+x²)-(2/3)∫x²dx+(2/3)∫[1/(1+x²)]dx-(2/3)∫dx
=(1/3)x³ln(1+x²)-(2/3)(x³/3)-(2/3)arctanx-(2/3)x+C
=(1/3)x³ln(1+x²)-(2/9)x³-(2/3)arctanx-(2/3)x+C