Question

integration of v.dv/1+v

Answer 1

v.dv/1+v=1+v-1.dv/1+v=(.dv)-(.dv/1+v)=v-log(1+v)

Answer 2

take the upper V as (1+V-1)
so V/(1+V)=(1+V-1)/(1+V)=(1+V)/(1+V)-1/(1+V)
=1-1/(1+V)
now the integration will be in the form
integration of [1-1/(1+V)]dV
= integration of [dV-dV/(1+V)]
=integration of dV-integration of dV/(1+V)
=V-integration of dV/(1+V).... (1)

now [let 1+V=u so dV=du
so integration of dV/(1+V)
=integration of du/u=log u=log (1+V)]
so from (1) we get
=V-log (1+V)
ans