∫x5m−1+2x4m−1(x2m+xm+1)3dx=f(x)+c thenf(x)=

∫x5m−1+2x4m−1x6m(1+x−m+x−2m)3=∫x−m−1+2x−2m−1(1+x−m+x−2m)3dx put    t=1+x−m+x−2m​​​​​​​​​       dtdx=−mx−m−1−2mx−2m−1    −dtm=(x−m−1+2x−2m−1)dx ∴∫x5m−1+2x4m−1(x2m+xm+1)3=−1m∫t−3dt=12mt2+c =12m(1+x−m+x−2m)2+c =x4m2m(x2m+xm+1)2+c