If n∈N and In=∫logxndx, thenIn+nIn-1=
logxn+1n+1
xlogxn+c
logxn-1
logxnn
Explanation for the correct option
Given that In=∫logxndx
∴In=∫logxn.1dx=logxn∫1dx-∫ddxlogxn∫1dxdx∵∫u.v=u.∫vdx-∫dudx∫vdxdx=xlogxn-∫nlogxn-1.1x.xdx∵ddxfgx=f'gx.g'x;ddxlogx=1x=xlogxn-∫nlogxn-1dx⇒In=xlogxn-nIn-1+C∵In=∫logxndx⇒In+nIn-1=xlogxn+C
Hence, the correct option is option(B) i.e. xlogxn+c