∫(ex)(x-1)x2dx=
exx2+c
-exx2+c
exx+c
-exx+c
Explanation for the correct option:
Given, ∫(ex)(x-1)x2dx
=∫(ex)1x-1x2dx
=exx+c [∵∫ex[f(x)+f'(x)]dx=exf(x)+C]
Hence, Option ‘C’ is Correct.
The general solution of a differential equation of the type dxdy+P1x=Q1 is
(a) ye∫P1 dy=∫(Q1e∫P1 dy)dy+C
(b) ye∫P1 dx=∫(Q1e∫P1 dx)dx+C
(c) xe∫P1 dy=∫(Q1e∫P1 dy)dy+C
(d) xe∫P1 dx=∫(Q1e∫P1 dx)dx+C