∫x5m−1+2x4m−1x2m+xm+13dx is equal to where m∈R−{0}

I=∫x5m−1+2x4m−1dxx6m1+x−m+x−2m3=∫x−(m+1)+2⋅x−(2m+1)1+x−m+x−2m3dx Put 1+x−m+x−2m=t-mx-m-1-2mx-2m-1dx=dt∴ I=−1m∫dtt3=-1mt-3+1-3+1=12mt2+c=x4m2mx2m+xm+12+c