Question

Задание приложено...

Answer 1

Ответ:

1.9

$\boxed{ \boldsymbol{\displaystyle I = \int {\frac{dx}{9x^{2} + 16} } = \frac{1}{12} \ \text{arctg} \bigg( \frac{3x}{4} \bigg) + C }}$

1.10

$\boxed{ \boldsymbol{\displaystyle I = \int {\frac{3^{x - 1} - 2^{x + 1}}{6^{x}} } \, dx = \frac{2 \cdot 3^{-x}}{\ln 3} - \frac{2^{-x}}{3\ln 2} + C}}$

Примечание:

По таблице интегралов:

$\boxed{\int {a^{x}} \, dx = \frac{a^{x}}{\ln a} + C }$

$\boxed{\int {\frac{dx}{x^{2} + a^{2}} } = \frac{1}{a} \ \text{arctg} \ \frac{x}{a} + C }$

По свойствам интегралов:

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

Пошаговое объяснение:

1.9

$\displaystyle I = \int {\frac{dx}{9x^{2} + 16} } = \int {\frac{dx}{(3x)^{2} + 4^{2}} } = \frac{1}{3} \int {\frac{d(3x)}{(3x)^{2} + 4^{2}} } = \frac{1}{12} \ \text{arctg} \bigg( \frac{3x}{4} \bigg) + C$

1.10

$\displaystyle I = \int {\frac{3^{x - 1} - 2^{x + 1}}{6^{x}} } \, dx = \int {\frac{3^{-1} \cdot 3^{x} - 2 \cdot2^{x}}{(2 \cdot 3)^{x}} } \, dx = \int {\frac{3^{-1} \cdot 3^{x} - 2 \cdot2^{x}}{2^{x} \cdot 3^{x}} } \, dx =$

$\displaystyle = \int {\frac{3^{-1} \cdot 3^{x} - 2 \cdot2^{x}}{2^{x} \cdot 3^{x}} } \, dx = \int \bigg( {\frac{3^{-1} \cdot 3^{x}}{2^{x} \cdot 3^{x}} - \frac{2 \cdot2^{x}}{2^{x} \cdot 3^{x}} } \bigg) \, dx = \int \bigg( {\frac{3^{-1}}{2^{x} } - \frac{2 }{3^{x}} } \bigg) \, dx =$

$\displaystyle = \int { \frac{3^{-1}}{2^{x} } } \, dx - \int { \frac{2 }{3^{x}} } \, dx = \frac{1}{3} \int {2^{-x}} \, dx - 2 \int {3^{-x}} \, dx =$

$\displaystyle = -\frac{1}{3} \int {2^{-x}} \, d(-x) - 2 \int {-3^{-x}} \, d(-x) = 2 \int {3^{-x}} \, d(-x) -\frac{1}{3} \int {2^{-x}} \, d(-x)=$

$\displaystyle = \frac{2 \cdot 3^{-x}}{\ln 3} + C_{1} - \frac{2^{-x}}{3\ln 2} + C_{2} = \frac{2 \cdot 3^{-x}}{\ln 3} - \frac{2^{-x}}{3\ln 2} + C$