∫dxx(log⁡x)2+4log⁡x+1

∫dxxlogx2+4logx+1 Put log⁡x=t1xdx=dt∫dxx(log⁡x)2+4log⁡x+1=∫dtt2+4t+1 Use ∫1x2−a2dx=12alog⁡x−ax+a+c=∫dtt2+2.2t+(2)2+1−(2)2=∫dt(t+2)2−(3)=∫dt(t+2)2−(3)2=123log⁡t+2−3t+2+3+c=123log⁡log⁡x+2−3log⁡x+2+3