∫1+x43xdx is equal to

∫1+x43xdx Let x=t4dx=4t3dt=∫1+t3t24t3dt=4∫1+t3tdt Put 1+t=k3dt=3k2dk=3k2dk=4∫kk3−13k2dk=12∫k3k3−1dk=12∫k6−k3dk=12k77−k44+C=127k7−124k4+C=127(1+t)73−3(1+t)43+C=1271+x473−31+x443+C