Ответ:

ответ на фото.....

Объяснение:

Ответ:

• 1.1 [tex]\boxed{ \boldsymbol{ I = \frac{2x^{3,5}}{7} - \frac{4x^{1,5}}{3} + 6\sqrt{x} + C } }[/tex]

• 1.2 [tex]\boxed{ \boldsymbol{I= -2\ \textbf{ctg}(2x) + C = \textbf{tg} \ x -\textbf{ctg} \ x + C } }[/tex]

• 1.3 [tex]\boxed{ \boldsymbol{ I = \frac{4}{3} \ln |3x - 1| + \dfrac{5\sin (2x + 4)}{2} + C } }[/tex]

Примечание:

По таблице интегралов:

[tex]\boxed{\displaystyle \int x^{n} \ dx = \frac{x^{n + 1}}{n + 1} + C; n \neq -1, x > 0}[/tex]

[tex]\boxed{\displaystyle \int \frac{1}{\sin^{2} x} \ dx = -\text{ctg}\ x + C}[/tex]

[tex]\boxed{\displaystyle \int \frac{1}{\cos^{2} x} \ dx = \text{tg}\ x + C}[/tex]

[tex]\boxed{\displaystyle \int \cos x \ dx = \sin x + C}[/tex]

[tex]\boxed{\displaystyle \int \frac{1}{x} \ dx = \ln|x| + C}[/tex]

По свойствам интегралов:

[tex]\boxed{ \displaystyle \int \sum\limits_{i=1}^n {C_{i}f_{i}(x)} \, dx = \sum\limits_{i=1}^nC_{i} \int {f_{i}(x)} \, dx}[/tex]

Объяснение:

1.1

[tex]\displaystyle I = \int { \frac{x^{3} - 2x + 3}{\sqrt{x} } } \, dx = \int { \frac{x^{3} - 2x + 3}{ x^{0,5} } } \, dx = \int { \bigg( \frac{x^{3} }{ x^{0,5} } - \frac{2x^{1} }{x^{0,5}} + \frac{3}{x^{0,5}} \bigg) } \, dx =[/tex]

[tex]\displaystyle = \int { \bigg( x^{3 - 0,5} - 2x^{1 -0,5}+ 3 \cdot x^{-0,5} \bigg) } \, dx = \int { \bigg( x^{2,5} - 2x^{0,5}+ 3 \cdot x^{-0,5} \bigg) } \, dx =[/tex]

[tex]\displaystyle= \int { x^{2,5} } \, dx - \int { 2x^{0,5} } \, dx + \int { 3x^{-0,5} } \, dx = \int { x^{2,5} } \, dx - 2\int { x^{0,5} } \, dx + 3\int { x^{-0,5} } \, dx =[/tex]

[tex]\displaystyle = \frac{x^{2,5 + 1}}{2,5 + 1} + C_{1} - \frac{2x^{0,5 + 1}}{0,5 + 1} + C_{2} + \frac{3x^{-0,5 + 1}}{-0,5 + 1} + C_{3} =[/tex]

[tex]\displaystyle = \frac{x^{3,5}}{3,5} - \frac{2x^{1,5}}{1,5 } + \frac{x^{0,5}}{0,5} + C = \frac{\dfrac{x^{3,5}}{1} }{\dfrac{35}{10} } - \frac{\dfrac{2x^{1,5}}{y} }{ \dfrac{15}{10} } + \frac{\dfrac{3x^{0,5}}{1} }{ \dfrac{5}{10} } + C =[/tex]

[tex]\displaystyle = \frac{10x^{3,5}}{35} - \frac{20x^{1,5}}{15 } + \frac{30x^{0,5}}{5} + C = \frac{2x^{3,5}}{7} - \frac{4x^{1,5}}{3} + 6\sqrt{x} + C[/tex]

1.2

[tex]\displaystyle I = \int {\frac{dx}{\sin^{2}x \cos^{2} x} } = -2\ \text{ctg}(2x) + C = \text{tg} \ x -\text{ctg} \ x + C[/tex]

1) способ решения

[tex]\displaystyle \int {\frac{dx}{\sin^{2}x \cos^{2} x} } = \int {\frac{4\ dx}{4\sin^{2}x \cos^{2} x} } = 4\int {\frac{ dx}{2 \sin x \cos x \cdot 2 \sin x \cos x} } =[/tex]

[tex]\displaystyle = 4\int {\frac{ dx}{\sin 2x \cdot \sin 2x } = 4\int {\frac{ dx}{\sin^{2} 2x } = \frac{4}{2} \int {\frac{ d(2x)}{\sin^{2} 2x } =2\int {\frac{ d(2x)}{\sin^{2} 2x } =[/tex]

[tex]= -2\ \text{ctg}(2x) + C[/tex]

2) способ решения

[tex]\displaystyle \int {\frac{dx}{\sin^{2}x \cos^{2} x} } = \int {\frac{\sin^{2} x + \cos^{2} x}{\sin^{2}x \cos^{2} x} } \, dx = \int {\bigg( \frac{\sin^{2} x }{\sin^{2}x \cos^{2} x} + \frac{ \cos^{2} x}{\sin^{2}x \cos^{2} x} \bigg) } \, dx =[/tex]

[tex]\displaystyle =\int {\bigg( \frac{1 }{ \cos^{2} x} + \frac{ 1}{\sin^{2}x } \bigg) } \, dx = \int { \frac{1 }{ \cos^{2} x} } \, dx + \int { \frac{ 1}{\sin^{2}x } } \, dx=[/tex]

[tex]= \text{tg} \ x + C_{1} -\text{ctg} \ x + C_{2}= \text{tg} \ x -\text{ctg} \ x + C[/tex]

1.3

[tex]\displaystyle I = \int {\bigg( \frac{4}{3x - 1} + 5 \cos(2x +4) \bigg)} \, dx = \int { \frac{4}{3x - 1} } \, dx + \int { 5 \cos(2x +4) } \, dx =[/tex]

[tex]\displaystyle = 4 \int { \frac{1}{ 3x - 1 } } \, dx + 5 \int { \cos(2x +4) } \, dx =[/tex]

[tex]\displaystyle =\dfrac{4}{3} \int { \frac{d(3x - 1)}{3x - 1 } } + \frac{5}{2} \int { \cos(2x +4) } \, d(2x + 4) =[/tex]

[tex]\displaystyle = \frac{4}{3} \ln |3x - 1 | + C_{1} + \dfrac{5\sin (2x + 4)}{2} + C_{2} = \frac{4}{3} \ln |3x - 1| + \dfrac{5\sin (2x + 4)}{2} + C[/tex]