Question

Нужно решить все задания

Answer 1

Ответ:

1.

при а = 4, b = 2

2.

F(x) = e^x

G(x) = e^x + 2

3.

$\int\limits(3x + x \sqrt[3]{x} )dx = \int\limits(3x + {x}^{ \frac{4}{3} } )dx = \\ = \frac{3 {x}^{2} }{2} + \frac{ {x}^{ \frac{7}{3} } }{ \frac{7}{3} } + c = \frac{3 {x}^{2} }{2} + \frac{3}{7} {x}^{2} \sqrt[3]{x} + c$

4.

$\int\limits \frac{ {x}^{2} + 2}{ {x}^{2} + 1} dx = \int\limits \frac{ {x}^{2} + 1 + 1 }{ {x}^{2} + 1} dx = \int\limits \: dx + \int\limits \frac{dx}{ {x}^{2} + 1} = x + arctg(x) + c$

5.

$\int\limits \frac{dx}{ { \cos }^{2} (4x)} = \frac{1}{4} \int\limits \frac{d(4x)}{ { \cos}^{2} (4x)} = \frac{1}{4} tg(4x) + c$

6.

$\int\limits {e}^{8x + 3} dx = \frac{1}{8} \int\limits {e}^{8x + 3} d(8x + 3) = \frac{1}{8} {e}^{8x + 3} + c$

7.

$\int\limits \frac{3dx}{6x + 1} = \frac{1}{2} ln(6x + 1) + c$

при a = 2, b= 6.

8.

$\int\limits \frac{dx}{ {(1 - 2x)}^{2} } = - \frac{1}{2} \times \frac{ {(1 - 2x)}^{ - 1} }{ - 1} + c = \frac{1}{2} \times {(1 - 2x)}^{ - 1} + c$

при a = 2, b = -1

9.

$\int\limits \: x {e}^{3x} dx$

$U = x \: \: \: \: \: \: \: \: \: \: \: \: \: \: dU = dx \\ dV = {e}^{3x} dx \: \: \: \: \: \: \: \: \: \: V = \frac{1}{3} {e}^{3x}$

$\frac{x}{3} {e}^{3x} - \frac{1}{3} \int\limits {e}^{3x} dx = \\ = \frac{x}{3} {e}^{3x} - \frac{1}{9} {e}^{3x} + c = \\ = \frac{ {e}^{3x} }{3} (x - \frac{1}{3} ) + c$

10.

$\int\limits(2x + 3) \sin( {x}^{2} + 3x - 1 ) dx$

${x}^{2} + 3x - 1 = t \\ (2x + 3)dx = dt$

$\int\limits \sin(t) dt = - \cos(t) + c = \\ = - \cos( {x}^{2} + 3x - 1) + c$