Calculus Question - need help?

Question

Consider the integral 3x(x^2+1). In the following, we will evaluate the integral using two methods.

Answer 1

ʃ[3x(x²+1)]dx

Distribute terms using distributive property:  a( b + c) = ab + ac

ʃ[3x³+3x] dx

So the original expression equals  ʃ[3x³+3x] dx

Split into two integrals ʃ3x³ dx + ʃ3x dx

Cavalier's quadrate formula: ʃx^a =(x^a+1) /(a+1)+C.

ʃ3x³ dx = (3x⁴)/4+C

ʃ3x dx = (3x²)/2+C

So the original integral is equal to (3/4)x⁴ + (3/2)x² +C.