91Ó°ÊÓ

Integrate the function \(\frac{x}{x+y}\) with respect to y. Treat x as a constant while integrating with respect to y. \(\int_{1}^{2} \frac{x}{x+y} dy\) Using substitution method let \(u = x + y, du = dy\) So the integral becomes, \(\int \frac{x}{u} du\) Now the limits of the integral also need to change according to the substitution: When \(y = 1, u = x+1\) When \(y = 2, u = x+2\) Now, integrate with respect to u: \(x \int_{x+1}^{x+2} \frac{1}{u} du = x[\ln(u)|_{x+1}^{x+2}] = x[\ln(x+2) - \ln(x+1)]\)

Integrate the result from Step 1 with respect to x, from 1 to 2. \(\int_{1}^{2} x[\ln(x+2) - \ln(x+1)] dx\) Now apply integration by parts, let \(u = x\), \(du = dx\) \(dv = [\ln(x+2) - \ln(x+1)] dx\), \(v = \int[\ln(x+2) - \ln(x+1)] dx\) To find v, we need to integrate \(\ln(x+2) dx - \int \ln(x+1) dx\) separately: \(v = \int\ln(x+2) dx - \int\ln(x+1) dx\) Use integration by parts for each integral: For \(\int\ln(x+2) dx\): \(w= \ln(x+2)\), \(dw =\frac{1}{x+2} dx\) \(dw_1=dx\), \(w_1 = x\) \(v_1 = x\ln(x+2) - \int x\frac{1}{x+2} dx\) For \(\int \ln(x+1) dx\): \(w = \ln(x+1)\), \(dw=\frac{1}{x+1} dx\) \(dw_2 = dx\), \(w_2 = x\) \(v_2 = x\ln(x+1) - \int x\frac{1}{x+1} dx\) So, \(v = v_1 - v_2 = x\ln(x+2) - \int x\frac{1}{x+2} dx - x\ln(x+1) + \int x\frac{1}{x+1} dx\) Now calculate, \(\int_{1}^{2} x[\ln(x+2) - \ln(x+1)] dx = \int_{1}^{2} [x\ln(x+2) - x\ln(x+1)] dx\) Using the integration by parts result we have, \(= [x\ln(x+2) - \int x\frac{1}{x+2} dx]_{1}^{2} + [\int x\frac{1}{x+1} dx - x\ln(x+1)]_{1}^{2} \) After evaluating all these terms for the limits 1 and 2, the final answer will be: $$\ln\left(\frac{5}{3}\right) + 1$$