Question

Выполнить примеры.
тт​

Answer 1

Ответ:

1.

$\int\limits {(x - 2)}^{3} dx = \int\limits{(x - 2)}^{3} d(x - 2) = \\ = \frac{ {(x - 2)}^{4} }{4} + C$

2.

$\int\limits \frac{ {x}^{5} + 3 {x}^{4} + 6 {x}^{3} }{ {x}^{2} } dx =\int\limits ( {x}^{3} + 3 {x}^{2} + 6x) dx = \\ = ( \frac{ {x}^{4} }{4} + 3 \times \frac{ {x}^{3} }{3} + \frac{6 {x}^{2} }{2} + C= \frac{ {x}^{4} }{4} + {x}^{3} + 3 {x}^{2} + C$

3.

$\int\limits \frac{ \sqrt{x} + \sqrt[3]{ {x}^{2} } + \sqrt[4]{ {x}^{3} } }{ {x}^{2} } dx = \int\limits \frac{ {x}^{ \frac{1}{2} } + {x}^{ \frac{2}{3} } + {x}^{ \frac{3}{4} } }{ {x}^{2} } dx = \\ = \int\limits( {x}^{ \frac{1}{2} - 2 } + {x}^{ \frac{2}{3} - 2} + {x}^{ \frac{3}{4} - 2 }) dx =\int\limits( {x}^{ - \frac{3}{2} } + {x}^{ - \frac{4}{3} } + {x}^{ - \frac{5}{3} } ) dx = \\ = \frac{ {x}^{ - \frac{1}{2} } }{ - \frac{1}{2} } + \frac{ {x}^{ - \frac{1}{3} } }{( - \frac{1}{3}) } + \frac{ {x}^{ - \frac{2}{3} } }{( - \frac{2}{3} )} + C= \\ = - \frac{2}{ \sqrt{x} } - \frac{3}{ \sqrt[3]{x} } - \frac{3}{2 \sqrt[3]{ {x}^{2} } } + C$

Answer 2

Ответ:

$\int (x-2)^3\, dx=[\ t=x-2\ ,\ dx=dt\ ]=\int t^3\, dt=\dfrac{t^4}{4}+C=\dfrac{(x-2)^4}{4}+C$

$2)\ \ \int \dfrac{x^5+3x^4+6x^3}{x^2}\, dx=\int (x^3+3x^2+6x)\, dx=\dfrac{x^4}{4}+x^3+3x^2+C\\\\\\3)\ \ \int \dfrac{\sqrt{x}+\sqrt[3]{x^2}+\sqrt[4]{x^3}}{x^2}\, dx=\int (x^{-\frac{3}{2}}+x^{-\frac{4}{3}}+x^{-\frac{5}{4}})\, dx=\\\\=\dfrac{x^{-\frac{1}{2}}}{-\frac{1}{2}}+\dfrac{x^{-\frac{1}{3}}}{-\frac{1}{3}}+\dfrac{x^{\frac{1}{4}}}{-\frac{1}{4}}+C=-\dfrac{2}{\sqrt{x} }-\dfrac{3}{\sqrt[3]{x}}-\dfrac{4}{\sqrt[4]{x}} +C$