Evaluate the iterated integral. \int_{1}^{3}\int_{0}^{y}\frac{4}{x^{2}+y^{2}}dx dy

Evaluate the iterated integral.
$$\displaystyle{\int_{{{1}}}^{{{3}}}}{\int_{{{0}}}^{{{y}}}}{\frac{{{4}}}{{{x}^{{{2}}}+{y}^{{{2}}}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$

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Step 1
Consider the following integral:
$$\displaystyle{\int_{{{1}}}^{{{3}}}}{\int_{{{0}}}^{{{y}}}}{\frac{{{4}}}{{{x}^{{{2}}}+{y}^{{{2}}}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={I}$$
Consider the following formula:
$$\displaystyle\int{\frac{{{1}}}{{{x}^{{{2}}}+{a}^{{{2}}}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{a}}}}{\arctan{{\left({\frac{{{x}}}{{{a}}}}\right)}}}+{C}$$
$$\displaystyle{I}={4}{\int_{{{1}}}^{{{3}}}}{{\left[{\frac{{{1}}}{{{y}}}}{\arctan{{\left({\frac{{{x}}}{{{y}}}}\right)}}}\right]}_{{{0}}}^{{{y}}}}{\left.{d}{y}\right.}$$
$$\displaystyle={4}{\int_{{{1}}}^{{{3}}}}{\left[{\frac{{{1}}}{{{y}}}}{\arctan{{\left({\frac{{{y}}}{{{y}}}}\right)}}}-{\frac{{{1}}}{{{y}}}}{\arctan{{\left({\frac{{{0}}}{{{y}}}}\right)}}}\right]}{\left.{d}{y}\right.}$$
$$\displaystyle={4}{\int_{{{1}}}^{{{3}}}}{\frac{{\pi}}{{{4}{y}}}}{\left.{d}{y}\right.}$$
$$\displaystyle={\int_{{{1}}}^{{{3}}}}{\frac{{\pi}}{{{y}}}}{\left.{d}{y}\right.}$$
Step 2
Consider the following formula:
$$\displaystyle\int{\frac{{{1}}}{{{y}}}}{\left.{d}{y}\right.}={\ln}{\left|{y}\right|}+{C}$$
$$\displaystyle{I}=\pi{{\left[{\ln}{\left|{y}\right|}\right]}_{{{1}}}^{{{3}}}}$$
$$\displaystyle=\pi{\left[{\ln{{\left({3}\right)}}}-{\ln{{\left({1}\right)}}}\right]}$$
$$\displaystyle=\pi{\ln{{\left({3}\right)}}}$$
Step 3
Hence, the solution is $$\displaystyle\pi{\ln{{\left({3}\right)}}}$$.