!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!помогите найти неопределенный интеграл:
1)знак интеграла (3x-2)^2dx
2)знак интеграла (2-x)^4-17x^9+корень из 2)dx
Ответы
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1) I. ![int {(3x-2)^2} , dx = int {9x^2-12x+4} , dx = 9int {x^24} , dx - 12int {x} , dx +\+ 4int {} , dx = 9cdot frac{x^3}{3} - 12cdotfrac{x^2}{2} + 4x = 3x^3 - 6x^2 + 4x + C, int {(3x-2)^2} , dx = int {9x^2-12x+4} , dx = 9int {x^24} , dx - 12int {x} , dx +\+ 4int {} , dx = 9cdot frac{x^3}{3} - 12cdotfrac{x^2}{2} + 4x = 3x^3 - 6x^2 + 4x + C,](https://tex.z-dn.net/?f=int+%7B%283x-2%29%5E2%7D+%2C+dx+%3D+int+%7B9x%5E2-12x%2B4%7D+%2C+dx+%3D+9int+%7Bx%5E24%7D+%2C+dx+-+12int+%7Bx%7D+%2C+dx+%2B%5C%2B+4int+%7B%7D+%2C+dx+%3D+9cdot+frac%7Bx%5E3%7D%7B3%7D+-+12cdotfrac%7Bx%5E2%7D%7B2%7D+%2B+4x+%3D+3x%5E3+-+6x%5E2+%2B+4x+%2B+C%2C+)
II.![int {(3x-2)^2} , dx = frac{1}{3}int {(3x-2)^2} , d(3x-2) = frac{1}{9}(3x-2)^3+C int {(3x-2)^2} , dx = frac{1}{3}int {(3x-2)^2} , d(3x-2) = frac{1}{9}(3x-2)^3+C](https://tex.z-dn.net/?f=int+%7B%283x-2%29%5E2%7D+%2C+dx+%3D+frac%7B1%7D%7B3%7Dint+%7B%283x-2%29%5E2%7D+%2C+d%283x-2%29+%3D+frac%7B1%7D%7B9%7D%283x-2%29%5E3%2BC)
2)![int {(2-x)^4-17x^9+sqrt{2}} , dx = -int {(2-x)^4} , d(2-x) - 17int {x^9} , dx +\+ sqrt{2}int {} , dx = -frac{1}{5}(2-x)^5 - frac{17}{10} x^{10} + x sqrt{2} = 0,2(x-2)^5 - 1,7x^{10} + x sqrt{2}+C int {(2-x)^4-17x^9+sqrt{2}} , dx = -int {(2-x)^4} , d(2-x) - 17int {x^9} , dx +\+ sqrt{2}int {} , dx = -frac{1}{5}(2-x)^5 - frac{17}{10} x^{10} + x sqrt{2} = 0,2(x-2)^5 - 1,7x^{10} + x sqrt{2}+C](https://tex.z-dn.net/?f=+int+%7B%282-x%29%5E4-17x%5E9%2Bsqrt%7B2%7D%7D+%2C+dx+%3D++-int+%7B%282-x%29%5E4%7D+%2C+d%282-x%29+-+17int+%7Bx%5E9%7D+%2C+dx+%2B%5C%2B+sqrt%7B2%7Dint+%7B%7D+%2C+dx+%3D+-frac%7B1%7D%7B5%7D%282-x%29%5E5+-+frac%7B17%7D%7B10%7D+x%5E%7B10%7D+%2B+x+sqrt%7B2%7D+%3D+0%2C2%28x-2%29%5E5+-+1%2C7x%5E%7B10%7D+%2B+x+sqrt%7B2%7D%2BC)
II.
2)
Ответ дал:
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