Begin by noting that since #15# is constant, we can pull it out of the integral, simplifying the problem down to:
#15int_-1^1x^2(1+x^3)^4dx#
Also note that the derivative of #x^3# is #x^2#, and we have an #x^2# in the integral. That makes this problem a textbook candidate for a #u#-substitution:
Let #color(blue)u=color(blue)(1+x^3)#
#u=1+(-1)^3=color(green)0-># for the boundaries of the integral
#u=1+(1)^3=color(green)2-># for the boundaries of the integral
#(du)/dx=3x^2#
#color(red)(du)=color(red)(3x^2dx)#
Using the substitution,
#5int_color(green)0^color(green)2color(red)(3x^2)(color(blue)(1+x^3))^4color(red)dx=5int_color(green)0^color(green)2color(blue)u^4color(red)(du)#
And this integral is a simple case of reverse power rule:
#5int_0^2u^4du=[5(u^5/5)]_0^2=[u^5]_0^2#
#color(white)(XX)=((2)^5-(0)^5)=32#
