91Ó°ÊÓ

First, we need to integrate the given integrand \(\frac{x}{x+y}\) with respect to y: $$\int_{1}^{2} \left[\int_{1}^{2} \frac{x}{x+y} d y\right] d x$$ Since x is constant while integrating with respect to y, we can treat it as a constant: $$\int_{1}^{2} \left[x \int_{1}^{2} \frac{1}{x+y} d y\right] d x$$ Now, integrate \(\frac{1}{x+y}\) with respect to y: $$\int \frac{1}{x+y} dy = \ln|x+y|$$ So we have: $$\int_{1}^{2} \left[x \left(\ln|x+y|\right)\right]_{1}^{2} d x$$

Next, we will evaluate the integral within the limits for y: $$\int_{1}^{2} \left[x (\ln|(x+2)| - \ln{(x+1)})\right] d x$$ We can simplify the expression using logarithmic properties: $$\int_{1}^{2} x\left(\ln{\frac{x+2}{x+1}}\right) d x$$

Now, we will integrate with respect to x: $$\int x\ln\frac{x+2}{x+1} dx$$ We can use integration by parts here: let \(u = \ln{\frac{x+2}{x+1}}\) and \(dv = x dx\). Then, we have \(du = \frac{1}{(x+2)(x+1)}dx\) and \(v = \frac{1}{2}x^2\). The integration by parts formula is: $$\int udv = uv - \int vdu$$ Now, apply the integration by parts formula: $$\int x\ln\frac{x+2}{x+1} dx = \frac{1}{2}x^2\ln\frac{x+2}{x+1} - \int \frac{1}{2}x^2 \frac{1}{(x+2)(x+1)}dx$$ To evaluate the remaining integral, we can use partial fractions. The integral then becomes: $$\frac{1}{2}x^2\ln\frac{x+2}{x+1} - \frac{1}{2}\int \left(\frac{A}{x+1} + \frac{B}{x+2}\right)x^2 dx$$ Solving the partial fractions equation, we obtain: $$\frac{1}{2}x^2\ln\frac{x+2}{x+1} - \frac{1}{2}\int \left(\frac{1}{x+1} - \frac{1}{x+2}\right)x^2 dx$$ Now, we can split the integral and integrate the separated terms: $$\frac{1}{2}x^2\ln\frac{x+2}{x+1} - \frac{1}{2} \left[\int \frac{x^2}{x+1} dx - \int \frac{x^2}{x+2} dx\right]$$ The two integrals on the right can be solved using substitution or using standard integral tables: $$\frac{1}{2}x^2\ln\frac{x+2}{x+1} - \frac{1}{2} \left[-\frac{x^2}{2} + x + \ln|x+1|- \left(-\frac{x^2}{2} + x - 2 + \ln|x+2| \right)\right]$$ Simplifying, we get: $$\frac{1}{2}x^2\ln\frac{x+2}{x+1} - \frac{1}{2} \left[-x^2 + 2x + \ln\frac{x+2}{x+1}\right]$$

Finally, we will evaluate the expression within the limits for x: $$\left[\frac{1}{2}x^2\ln\frac{x+2}{x+1} - \frac{1}{2} \left[-x^2 + 2x + \ln\frac{x+2}{x+1}\right]\right]_{1}^{2}$$ $$= \left(\frac{1}{2}(2^2)\ln\frac{4}{3} - \frac{1}{2} \left[-(2^2) + 2(2) + \ln\frac{4}{3}\right]\right) - \left(\frac{1}{2}(1^2)\ln\frac{3}{2} - \frac{1}{2} \left[-(1^2) + 2(1) + \ln\frac{3}{2}\right]\right)$$ Evaluating and simplifying the expression, we get: $$-\frac{1}{2} + \ln\frac{4}{3}$$ Thus, the value of the given iterated integral is: $$\int_{1}^{2} \int_{1}^{2} \frac{x}{x+y} d y d x = -\frac{1}{2} + \ln\frac{4}{3}$$