Question #320387

∫ 3𝑥 + 6 √2𝑥 2 + 8𝑥 + 3 𝑑x


1
Expert's answer
2022-04-03T15:13:24-0400

(3x3+62x2+8x+3)dx==3x44+62x33+8x22+3x+C==3x44+22x3+4x4+3x+C\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


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