Answer to Question #320387 in Calculus for koi

Question #320387

∫ 3π‘₯ + 6 √2π‘₯ 2 + 8π‘₯ + 3 𝑑x


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

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


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