Question

设f(x)的一个原函数为x^2lnx,求不定积分xf(x)dx,

Answer 1

∫f(x)=x²lnx

f(x)=lnx*2x+x²*1/x

=2xlnx+x

∫xf(x) dx

=∫x*(2xlnx+x) dx

=2∫lnx d(x³/3) + ∫x² dx

=(2/3)x³lnx - (2/3)∫x² dx + ∫x² dx,分部积分法

=(2/3)x³lnx + (1-2/3)*x³/3 + C

=(2/3)x³lnx + (1/9)x³ + C

=(1/9)x³(1+6lnx) + C