Integral of 2x+5+(2xy/(x^2+y^2)) dy

1. Integrate term-by-term:

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \frac{2 x y}{x^{2} + y^{2}}\, dy = 2 x \int \frac{y}{x^{2} + y^{2}}\, dy$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{y}{x^{2} + y^{2}}\, dy = \frac{\int \frac{2 y}{x^{2} + y^{2}}\, dy}{2}$$

         1. Let $u = x^{2} + y^{2}$.

            Then let $du = 2 y dy$ and substitute $\frac{du}{2}$:

            $$\int \frac{1}{2 u}\, du$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            Now substitute $u$ back in:

            $$\log{\left(x^{2} + y^{2} \right)}$$

         So, the result is: $\frac{\log{\left(x^{2} + y^{2} \right)}}{2}$

      So, the result is: $x \log{\left(x^{2} + y^{2} \right)}$

   1. The integral of a constant is the constant times the variable of integration:

      $$\int 2 x\, dy = 2 x y$$

   1. The integral of a constant is the constant times the variable of integration:

      $$\int 5\, dy = 5 y$$

   The result is: $2 x y + x \log{\left(x^{2} + y^{2} \right)} + 5 y$

2. Add the constant of integration:

   $$2 x y + x \log{\left(x^{2} + y^{2} \right)} + 5 y+ \mathrm{constant}$$

The answer is:

$$2 x y + x \log{\left(x^{2} + y^{2} \right)} + 5 y+ \mathrm{constant}$$