∫1x+x5dx=f(x)+c, then the value of ∫x4x+x5dx=
logx-f(x)+c
f(x)+logx+c
f(x)-logx+c
None of these
∫x4x+x5dx add & subtract 1 in the numerator= ∫x4+1x+x5dx− ∫1x+x5dx =∫1+x4x1+x4dx−∫1x+x5dx=∫1xdx−∫1x+x5dx =logx−fx+c given ∫1x+x5dx=fx