I'm assuming that you meant differentiate, because we cannot integrate using the product rule.
#f(x)=x(x^7+15)^3#
Use the product rule, which states that
#d/dx(uv)=u'v+uv'#
Let #u=x#, #v=(x^7+15)^3#
We now need to differentiate #v=(x^7+15)^3#, so we are going to use the chain rule, which states that
#dy/dx=dy/(du)*(du)/dx#
Since we already have a #u# for the product rule, and we are trying to differentiate #v#, let's rewrite the rule as
#(dv)/dx=(dv)/(da)*(da)/(dx)#
Let #a=x^7+15#, so #(da)/dx=7x^6#, #v=a^3#, #(dv)/(da)=3a^2#
Combining, we get #(dv)/dx=7x^6*3a^2#
#(dv)/dx=21x^6a^2#
Putting back #a=x^7+15#, we get
#(dv)/dx=21x^6(x^7+15)^2#
#v'=21x^6(x^7+15)^2#
Now that we have differentiated #v#, let's put it in the product rule
#d/dx(f(x))=1*(x^7+15)^3+x(21x^6(x^7+15)^2)#
#f'(x)=(x^7+15)^3+21x^7(x^7+15)^2#
