Задача 69575 ...

6355a2d83db1a168cec87f02

25.02.2023 17:35:29

вычислить определенный интеграл

∫ dx/(1+sqrt(3x-2))

5f3ea7e3faf909182968ddd9

26.02.2023 13:45:16

★

Вычисляем неопределенный интеграл:

Замена переменной:
[m]\sqrt{3x+2}=t[/m] ⇒ [m]3x+2=t^2[/m]

[m]x=\frac{1}{3}(t^2-2)[/m]

[m]dx=\frac{1}{3}\cdot 2tdt[/m]

[m]dx=\frac{2}{3}tdt[/m]

[m]∫ \frac{1}{1+\sqrt{3x+2}}dx= ∫ \frac{\frac{2}{3}tdt}{1+t}=\frac{2}{3} ∫ \frac{t+1-1}{1+t}dt=\frac{2}{3} ∫ (1-\frac{1}{1+t})dt=[/m]

[m]=\frac{2}{3}t-\frac{2}{3}ln|1+t|+C=\frac{2}{3}\sqrt{3x+2}-\frac{2}{3}ln(1+\sqrt{3x+2})+C[/m]


По формуле Ньютона - Лейбница:
[m] ∫ ^{6}_{1}\frac{1}{1+\sqrt{3x+2}}dx=(\frac{2}{3}\sqrt{3x+2}-\frac{2}{3}ln(1+\sqrt{3x+2}))|^{6}_{1}=[/m]

[m]=(\frac{2}{3}\sqrt{3\cdot 6+2}-\frac{2}{3}ln(1+\sqrt{3\cdot 6+2}))-(\frac{2}{3}\sqrt{3\cdot 1+2}-\frac{2}{3}ln(1+\sqrt{3\cdot 1+2}))=\frac{2}{3}\cdot 2\sqrt{5}-\frac{2}{3}ln(1+2\sqrt{5})-\frac{2}{3}\sqrt{5}+\frac{2}{3}ln(1+\sqrt{5})=[/m]

[m]=\frac{2}{3}\sqrt{5}-\frac{2}{3}ln\frac{1+\sqrt{5}}{1+2\sqrt{5}}[/m]

626fab9a354ab65fb2f43abb

25.02.2023 18:32:09

[m]\int_1^6 \frac{dx}{1+\sqrt{3x+2}}[/m]
Замена:
[m]\sqrt{3x+2} = y; dy = \frac{3dx}{2\sqrt{3x+2}} = \frac{3dx}{2y}; dx = \frac{2}{3}y\ dy[/m]
[m]y(1) = \sqrt{3+2} = \sqrt{5}; y(6) = \sqrt{3 \cdot 6+2} = \sqrt{20} = 2\sqrt{5}[/m]
Подставляем в интеграл:
[m]\frac{2}{3} \cdot \int_{\sqrt{5}}^{2\sqrt{5}} \frac{y\ dy}{1+y} = \frac{2}{3} \cdot \int_{\sqrt{5}}^{2\sqrt{5}} \frac{y+1-1}{y+1}dy =[/m]
[m]= \frac{2}{3} \cdot \int_{\sqrt{5}}^{2\sqrt{5}} (1 - \frac{1}{y+1})dy = \frac{2}{3} \cdot (y - ln(y+1))|_{\sqrt{5}}^{2\sqrt{5}} =[/m]
[m]= \frac{2}{3} \cdot (2\sqrt{5} - ln(2\sqrt{5}+1) - \sqrt{5} + ln(\sqrt{5}+1)) = [/m]
[m]= \frac{2}{3} \cdot (\sqrt{5} + ln \frac{\sqrt{5}+1}{2\sqrt{5}+1})[/m]