Evaluate ∫x2+4x4+16dx
122tan−1x2−4x2+C
122tan−1x−42x2+C
122tan−1x2−422+C
122tan−1x2−42x2+C
I=∫x2+4x4+16dx=∫1+4x2x2+16x2dx=∫1+4x2x2+4x2−8+8dx=∫1+4x2x−4x2+8dx Let x−4x=t . Then dx−4x=dt or 1+4x2dx=dt ∴ I=∫dtt2+(22)2=122tan−1t22+C=122tan−1x−4x22+C=122tan−1x2−42x2+C