How do I evaluate the indefinite integral of ∫ x^2 (√2+x) dx ?

Answer 1

Let u = 2 + x. Then du = dx and x = u - 2

So, we have ∫(u-2)² √udu = ∫ [u² - 4u + 4][u^1/2]du = ∫ [u^5/2 - 4u^3/2 + 4u^1/2]du

= (2/7)u^7/2 - (8/5)u^5/2 + (8/3)u^3/2 + C = (2/7)(x+2)^7/2 - (8/5)(x+2)^5/2 + (8/3)(x+2)^3/2 + C

Answer 2

As submitted the integral is [x³sqrt(2)/3] + x⁴/4....but it is not clear that you have submitted your question correctly or as you intended.