1) ∫ 2x³ - 3/x² + 7x⁶ dx =

Привет! Давай решим эти интегралы. Будем использовать формулу $\int x^n dx = \frac{x^{n+1}}{n+1} + C$. 1) $\int (2x^3 - 3x^{-2} + 7x^6) dx = 2 \cdot \frac{x^4}{4} - 3 \cdot \frac{x^{-1}}{-1} + 7 \cdot \frac{x^7}{7} + C = \frac{1}{2}x^4 + \frac{3}{x} + x^7 + C$ 2) $\int (3x+7)^2 dx = \int (9x^2 + 42x + 49) dx = 9 \cdot \frac{x^3}{3} + 42 \cdot \frac{x^2}{2} + 49x + C = 3x^3 + 21x^2 + 49x + C$ 3) $\int (4\sqrt{x} + 3\sqrt[3]{x}) dx = \int (4x^{1/2} + 3x^{1/3}) dx = 4 \cdot \frac{x^{3/2}}{3/2} + 3 \cdot \frac{x^{4/3}}{4/3} + C = \frac{8}{3}x^{3/2} + \frac{9}{4}x^{4/3} + C = \frac{8}{3}x\sqrt{x} + \frac{9}{4}x\sqrt[3]{x} + C$ 4) Разложим знаменатель $(x-1)(x^2+5x+6) = (x-1)(x+2)(x+3)$. Дробь $\frac{x^2-19x+6}{(x-1)(x+2)(x+3)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+3}$. Находим $A=-1, B=7, C=-5$. $\int (-\frac{1}{x-1} + \frac{7}{x+2} - \frac{5}{x+3}) dx = -\ln|x-1| + 7\ln|x+2| - 5\ln|x+3| + C$ 5) $\int \frac{dx}{x^2+4x+6} = \int \frac{dx}{(x+2)^2 + 2} = \frac{1}{\sqrt{2}}\arctan(\frac{x+2}{\sqrt{2}}) + C$ 6) $\int (2x + 3x^6 - 8x^3) dx = x^2 + \frac{3x^7}{7} - 2x^4 + C$ 7) $\int \sqrt{x}\sqrt[3]{x}\sqrt[4]{x} dx = \int x^{1/2+1/3+1/4} dx = \int x^{13/12} dx = \frac{x^{25/12}}{25/12} + C = \frac{12}{25}x^{25/12} + C$ 8) $\int \frac{dx}{\sqrt{9-x^2}} = \arcsin(\frac{x}{3}) + C$ 9) $\int \frac{1+2x^2}{x^2(1+x^2)} dx = \int (\frac{1}{x^2(1+x^2)} + \frac{2}{1+x^2}) dx$. Используем разложение $\frac{1}{x^2(1+x^2)} = \frac{1}{x^2} - \frac{1}{1+x^2}$. $\int (\frac{1}{x^2} - \frac{1}{1+x^2} + \frac{2}{1+x^2}) dx = \int (\frac{1}{x^2} + \frac{1}{1+x^2}) dx = -\frac{1}{x} + \arctan(x) + C$ 10) $\int \frac{1-x^2}{1+x^2} dx = \int (\frac{2-(1+x^2)}{1+x^2}) dx = \int (\frac{2}{1+x^2} - 1) dx = 2\arctan(x) - x + C$