∫ealogx+exlogadx=…..+c,a,x>1
xa+1(a+1)+axloga
ealogxalogx+exlogaxloga
xa-1a–1+ax.loga+c
ealogxax+exlogax
Explanation for the correct option:
Evaluating the integral:
Given
∫ealogx+exlogadx=∫(xa+exloga)dx∵enlogm=mn=xa+1(a+1)+exlogaloga+c=xa+1a+1+axloga+c
Hence, the correct answer is option (A).
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