# Calculate the double integral =\int \int_{R} \frac{4x}{1+xy}dA, R=[0, 4] \times [0,

Calculate the double integral
$$\displaystyle=\int\int_{{{R}}}{\frac{{{4}{x}}}{{{1}+{x}{y}}}}{d}{A},{R}={\left[{0},{4}\right]}\times{\left[{0},{1}\right]}$$

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Step 1
The given integrals is $$\displaystyle=\int\int_{{{R}}}{\frac{{{4}{x}}}{{{1}+{x}{y}}}}{d}{A}$$ wrehe
$$\displaystyle={R}=\frac{{{0},{4}}}{\times}{\left[{0},{1}\right]}$$
Evaluate as follows
Step 2
$$\displaystyle=\int\int_{{{R}}}{\frac{{{4}{x}}}{{{1}+{x}{y}}}}{d}{A}={\int_{{{0}}}^{{{1}}}}{\int_{{{0}}}^{{{4}}}}{\frac{{{4}{x}}}{{{1}+{x}{y}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle={4}{\int_{{{0}}}^{{{1}}}}{\int_{{{0}}}^{{{4}}}}{\left({\frac{{{1}}}{{{y}}}}-{\frac{{{1}}}{{{y}{\left({1}+{x}{y}\right)}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle={4}{\int_{{{0}}}^{{{1}}}}{{\left[{\frac{{{x}}}{{{y}}}}-{\frac{{{\ln{{\left({1}+{x}{y}\right)}}}}}{{{y}^{{{2}}}}}}\right]}_{{{0}}}^{{{4}}}}{\left.{d}{y}\right.}$$
$$\displaystyle={4}{\int_{{{0}}}^{{{1}}}}{\left({\frac{{{4}}}{{{y}}}}-{\frac{{{\ln{{\left({1}+{x}{y}\right)}}}}}{{{x}^{{{2}}}}}}\right)}{\left.{d}{y}\right.}$$
$$\displaystyle={4}{\int_{{{0}}}^{{{1}}}}{\frac{{{4}}}{{{y}^{{{2}}}}}}{\left({4}{y}-{\frac{{{\left\lbrace{\ln{{\left({1}+{4}{y}\right)}}}\right\rbrace}}}{{{4}}}}\right)}$$
$$\displaystyle={4}{\left[{5}{I}{n}{\left({5}\right)}-{4}\right]}$$ (by uv meth
$$\displaystyle={4}{\left({5}{I}{n}{\left({5}\right)}-{4}\right)}$$
Thus, $$\displaystyle=\int\int_{{{R}}}{\frac{{{4}{x}}}{{{1}+{x}{y}}}}{\left({5}{I}{n}{\left({5}\right)}-{4}\right)}$$.