The correct option is B 14ln∣∣∣lnx−xlnx+x∣∣∣−12tan−1(lnxx)+C
I=∫x2(1−lnx)x4((lnxx)4−1) dx
=∫1−lnxx2((lnxx)4−1) dx
Put lnxx=t⇒1−lnxx2=dt
∴I=∫1t4−1 dt
=∫dt(t2+1)(t2−1)
=12∫(t2+1)−(t2−1)(t2+1)(t2−1) dt
=12(∫dtt2−1−∫dtt2+1)
=12(12ln∣∣∣t−1t+1∣∣∣−tan−1t)+C
=14ln∣∣∣lnx−xlnx+x∣∣∣−12tan−1(lnxx)+C