\( \int \frac { x d x - y d y } { x ^ { 2 } + y ^ { 2 } } = \int \frac { 2 d z } { 1 - z ^ { 2 } } \)

Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int \frac{2}{1-z^{2}} dz\) - step1: Rewrite the expression: \(\int 2\times \frac{1}{1-z^{2}} dz\) - step2: Use properties of integrals: \(2\times \int \frac{1}{1-z^{2}} dz\) - step3: Evaluate the integral: \(2\times \frac{1}{2}\ln{\left(\left|\frac{z+1}{z-1}\right|\right)}\) - step4: Multiply the terms: \(\ln{\left(\left|\frac{z+1}{z-1}\right|\right)}\) - step5: Add the constant of integral C: \(\ln{\left(\left|\frac{z+1}{z-1}\right|\right)} + C, C \in \mathbb{R}\) Calculate the integral \( \int \frac{2 d z}{1 - z^2} \). Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int \frac{2}{1-z^{2}} dz\) - step1: Rewrite the expression: \(\int 2\times \frac{1}{1-z^{2}} dz\) - step2: Use properties of integrals: \(2\times \int \frac{1}{1-z^{2}} dz\) - step3: Evaluate the integral: \(2\times \frac{1}{2}\ln{\left(\left|\frac{z+1}{z-1}\right|\right)}\) - step4: Multiply the terms: \(\ln{\left(\left|\frac{z+1}{z-1}\right|\right)}\) - step5: Add the constant of integral C: \(\ln{\left(\left|\frac{z+1}{z-1}\right|\right)} + C, C \in \mathbb{R}\) Let's solve the integrals step by step. ### Step 1: Solve the left integral We need to evaluate the integral: \[ \int \frac{x \, dx - y \, dy}{x^2 + y^2} \] This integral can be separated into two parts: \[ \int \frac{x \, dx}{x^2 + y^2} - \int \frac{y \, dy}{x^2 + y^2} \] #### Part 1: \( \int \frac{x \, dx}{x^2 + y^2} \) Using the substitution \( u = x^2 + y^2 \), we have \( du = 2x \, dx \) or \( dx = \frac{du}{2x} \). The integral becomes: \[ \int \frac{x \, dx}{u} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln |u| + C_1 = \frac{1}{2} \ln |x^2 + y^2| + C_1 \] #### Part 2: \( \int \frac{y \, dy}{x^2 + y^2} \) Using the substitution \( v = x^2 + y^2 \), we have \( dv = 2y \, dy \) or \( dy = \frac{dv}{2y} \). The integral becomes: \[ \int \frac{y \, dy}{v} = \frac{1}{2} \int \frac{dv}{v} = \frac{1}{2} \ln |v| + C_2 = \frac{1}{2} \ln |x^2 + y^2| + C_2 \] ### Combining the results Thus, the left integral evaluates to: \[ \frac{1}{2} \ln |x^2 + y^2| - \frac{1}{2} \ln |x^2 + y^2| + C = C \] ### Step 2: Solve the right integral Now, we evaluate the integral: \[ \int \frac{2 \, dz}{1 - z^2} \] This integral can be solved using the formula for the integral of a rational function: \[ \int \frac{2 \, dz}{1 - z^2} = \ln \left| \frac{z + 1}{z - 1} \right| + C \] ### Final Result Putting it all together, we have: \[ C = \ln \left| \frac{z + 1}{z - 1} \right| + C \] Thus, the final result of the original equation is: \[ \int \frac{x \, dx - y \, dy}{x^2 + y^2} = \ln \left| \frac{z + 1}{z - 1} \right| + C \]