Question

ЗНАТОКИ АЛГЕБРЫ
Вычислите неопределенный интеграл (a,c,e)
​

Answer 1

Ответ:

$a)\ \ \int (2x^7+x^4-4x)\, dx=\dfrac{2x^8}{8}+\dfrac{x^5}{5}-\dfrac{4x^2}{2}+C=\dfrac{x^8}{4}+\dfrac{x^5}{5}-2x^2+C\\\\\\c)\ \ \int (\underbrace {x\cdot x^8}_{x^9}+x+x^8)\, dx=\dfrac{x^{10}}{10}+\dfrac{x^2}{2}+\dfrac{x^9}{9}+C\\\\\\e)\ \ \int (\, \underbrace {(x^5)^6}_{x^{30}}+3^{10}\cdot x^8-9^9\, )\, dx=\dfrac{x^{31}}{31}+3^{10}\cdot \dfrac{x^9}{9}+9^9\cdot x+C=\dfrac{x^{31}}{31}+3^{8}\cdot x^9+9^9\cdot x+C$

Answer 2

Ответ:

$\dfrac{x^{8}}{4}+\dfrac{x^{5}}{5}-2x^{2}+C, \quad C-const;$

$\dfrac{x^{10}}{10}+\dfrac{x^{2}}{2}+\dfrac{x^{9}}{9}+C, \quad C-const;$

$\dfrac{x^{31}}{31}+3^{8} \cdot x^{9}-9^{9} \cdot x+C, \quad C-const;$

Объяснение:

$\int\ {x^{n}} \, dx =\dfrac{x^{n+1}}{n+1}+C, \quad n \neq -1, \quad C-const;$

$\int\ {(2x^{7}+x^{4}-4x)} \, dx =2\int\ {x^{7}} \, dx +\int\ {x^{4}} \, dx -4\int\ {x} \, dx =2 \cdot \dfrac{x^{7+1}}{7+1}+\dfrac{x^{4+1}}{4+1}-4 \cdot$

$\cdot \dfrac{x^{1+1}}{1+1}+C=2 \cdot \dfrac{x^{8}}{8}+\dfrac{x^{5}}{5}-4 \cdot \dfrac{x^{2}}{2}+C=\dfrac{x^{8}}{4}+\dfrac{x^{5}}{5}-2x^{2}+C, \quad C-const;$

____________________________________________________________

$\int\ {(x \cdot x^{8}+x+x^{8})} \, dx =\int\ {x^{9}} \, dx +\int\ {x} \, dx +\int\ {x^{8}} \, dx =\dfrac{x^{9+1}}{9+1}+\dfrac{x^{1+1}}{1+1}+\dfrac{x^{8+1}}{8+1}+$

$+C=\dfrac{x^{10}}{10}+\dfrac{x^{2}}{2}+\dfrac{x^{9}}{9}+C, \quad C-const;$

____________________________________________________________

$\int\ {((x^{5})^{6}+3^{10} \cdot x^{8}-9^{9})} \, dx =\int\ {x^{30}} \, dx +3^{10}\int\ {x^{8}} \, dx -9^{9}\int\ {x^{0}} \, dx =\dfrac{x^{30+1}}{30+1}+$

$+3^{10} \cdot \dfrac{x^{8+1}}{8+1}-9^{9} \cdot \dfrac{x^{0+1}}{0+1}+C=\dfrac{x^{31}}{31}+3^{8} \cdot x^{9}-9^{9} \cdot x+C, \quad C-const;$