∫dxx(logx)2+4logx+1
123loglogx+2−3logx+2+3+c
123loglogx+2+3logx+2−3+c
123log2+3+logx2+3−logx+c
123log2+3−logx2+3+logx+c
∫dxxlogx2+4logx+1
Put logx=t
1xdx=dt∫dxx(logx)2+4logx+1=∫dtt2+4t+1
Use ∫1x2−a2dx=12alogx−ax+a+c
=∫dtt2+2.2t+(2)2+1−(2)2=∫dt(t+2)2−(3)=∫dt(t+2)2−(3)2=123logt+2−3t+2+3+c=123loglogx+2−3logx+2+3