∫1x+x5dx=f(x)+C, then the value of ∫x4x+x5dx is
logx-f(x)+C
f(x)+logx+C
f(x)-logx+C
None of these
Explanation for Correct answer:
Finding the integral value:
Given, ∫1x+x5dx=f(x)+C→(i)
∫x4x+x5dx=∫x4x1+x4dx=∫x4+1-1x1+x4dxaddandsubtract1innumerator=∫x4+1x1+x4-1x1+x4dx=∫1xdx-∫1x1+x4dx=logx-f(x)+Cfromequation(i)
Hence, option (A) is the correct answer.