This question aims to find the **double integral** of the given **expression** over a given **range** in $x-axi$ and $y-axis$.

This question is based on the concept of **integration, **particularly **double integrals.** The **integration** is used to find the **surface area** of **two-dimensional** regions and the **volume** of **three-dimensional** objects.

## Expert Answer

We have the following double integral expression given as:

\[ \iint_{R}^{} (\dfrac{6x}{1 + xy}) dA \]

The **range **is given as:

\[ R = {(x, y): 0 \le x \le 6, 0 \le y \le 1} \]

The following **formulas** are used to solve the question.

\[ \int x^n dx = \dfrac{x^{n + 1}}{n + 1} + C \]

\[ \int kx dx = k \dfrac{x^2}{2} + C \]

\[ \int \dfrac{1}{\sqrt{x}} dx = \int x^{-\frac{1}{2}} dx \]

\[ \iint_{R}^{} (\dfrac{6x}{1 + xy}) dA = \int_{0}^{6} \int_{0}^{1} \dfrac{6x}{1 + xy} dy dx \]

Based on the variables, we have separated the **integrals** for the $dx$ and $dy$ as:

\[ = \int_{0}^{6} 6x dx \int_{0}^{1} (1 + xy)^{-1} dy \]

\[ = \int_{0}^{6} 6x dx \left[ ln(1 +xy) \dfrac{1}{x} \right]_{0}^{1} \]

\[ = \int_{0}^{6} \dfrac{6x}{x} dx \left[ ln(1 +xy) \right]_{0}^{1} \]

By inserting the **integral values** and simplifying the expression as:

\[ = \int_{0}^{6} 6 dx \left[ln(1 + x) – 0 \right] \]

\[ = 6\int_{0}^{6} ln(1 + x) dx \]

\[ = 6\left[ln(1 + x)(1 + x) – x \right]_{0}^{6} \]

By inserting the **integral values** and simplifying the expression for $dy$ as:

\[ = 6\left[ln(1 + 6)(1 + 6) – 6 \right] \]

\[ = 42 \times ln(7) – 36 \]

\[ = 45.7 \]

## Numerical Results

The **double integral** of the given expression is as follows:

\[ \iint_{R} (\dfrac{6x}{1 + xy}) dA = 45.7 \]

## Example

Calculate the **double derivative** of the expression given below.

\[ \int_{1}^{2}\int_{4}^{9}\dfrac{3 + 5y}{\sqrt{x}} dx dy \]

Simplifying the expression:

\[ = \int_{1}^{2}\int_{4}^{9}(3 + 5y)x^{-\frac{1}{2}} dx dy \]

Then, based on the variables, we have separated the **integrals** for the $dx$ and $dy$ as:

\[ =\int_{1}^{2}(3 + 5y)dy \int_{4}^{9}x^{-\frac{1}{2}} dx \]

\[ = \int_{1}^{2}(3 + 5y)dy \left[ \frac{x^{- \frac{1}{2} + 1}}{\frac{-1}{2} + 1} \right]_{4}^{9} \]

\[ = \int_{1}^{2}(3 + 5y)dy \left[ \frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right]_{4}^{9} \]

We insert the **integral values** and simplify the expression for $dx$ as:

\[ = \int_{1}^{2}(3 + 5y)dy \left[ 2(9^{\frac{1}{2}} – 4^{\frac{1}{2}}) \right] \]

\[ = \int_{1}^{2}(3 + 5y)dy \left[ 2(3 – 2) \right] \]

\[ = 2\int_{1}^{2}(3 + 5y)dy \]

\[ = 2\left[3y + \frac{5y^2}{2} \right]_{1}^{2} \]

We insert the **integral values** and simplify the expression for $dy$ as follows:

\[ = 2\left[ 3(2 – 1) + \frac{5}{2}(2^2 – 1^2) \right] \]

\[ = 2\left[ 3 + 5 \times 1.5 \right] \]

\[ = 2(10.5) \]

\[ = 21 \]

Hence, we have the final value as: