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Answer to Question #320387 in Calculus for koi
Question #320387
β« 3π₯ + 6 β2π₯ 2 + 8π₯ + 3 πx
Expert's answer
"\\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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