If ∫x5m−1+2x4m−1x2m+xm+13dx=f(x)+C then f(x)=0

∫x5m−1+2x4m−1dxx6m1+1xm+1x2m3=∫x−m−1+2x−2m−1dx1+x−m+x−2m3=−1m∫dtt3 put t=1+x−m+x−2m=−1m1+x−m+x−2m−2+C=x4m2mx2m+xm+12+C