Integrate ∫logxdx.
Solve the given integral
Use integration by parts,
∫IxIIx=Ix∫IIxdx-∫dIxdx∫IIxdx
We know that, logx=lnx
So, for∫1×logx, where Ix=logx,IIx=1
∫logx=logx∫1dx-∫dlogxdx∫1dx∵∫cdx=cx,c=constant⇒∫logx=xlogx-∫1xxdx∵dlogxdx=1x⇒∫logx=xlogx-1+C
Hence, ∫logx is integrated as xlogx-1+C.