Answer to Question #320387 in Calculus for koi

Question

∫ 3𝑥 + 6 √2𝑥 2 + 8𝑥 + 3 𝑑x

Answer 1

$\int (3x^3 +6\sqrt{2} x^2 +8x + 3)dx =\\ =\cfrac{3x^4}{4}+ \cfrac{6\sqrt{2} x^3}{3} + \cfrac{8x^2}{2} + 3x+C = \\ =\cfrac{3x^4}{4} + 2\sqrt{2}x^3 + 4x^4 +3x+C$