$\int \frac{5 x^{4}+7 x^{2}+x+2}{x\left(x^{2}+1\right)^{2}} d x$

Step 1 ：\(\int \frac{5 x^{4}+7 x^{2}+x+2}{x\left(x^{2}+1\right)^{2}} d x = \int \frac{A}{x} + \frac{Bx + C}{x^2 + 1} + \frac{Dx + E}{(x^2 + 1)^2} dx\)

Step 2 ：\(5x^4 + 7x^2 + x + 2 = A(x^2 + 1)^2 + (Bx + C)x(x^2 + 1) + (Dx + E)x\)

Step 3 ：\(5x^4 + 7x^2 + x + 2 = Ax^4 + 2Ax^2 + A + Bx^3 + Bx + Cx^2 + C + Dx^3 + Dx + Ex^2 + E\)

Step 4 ：\(A = 5, B + D = 0, 2A + B + C + E = 7, C + E = 1, B + D = 1\)

Step 5 ：\(A = 5, B = -D, C = 1 - E, 2(5) - D + 1 - E + E = 7\)

Step 6 ：\(D = -3, B = 3, C = 1 - E, E = 1\)

Step 7 ：\(\int \frac{5 x^{4}+7 x^{2}+x+2}{x\left(x^{2}+1\right)^{2}} d x = \int \frac{5}{x} + \frac{3x - 3}{x^2 + 1} + \frac{-3x + 1}{(x^2 + 1)^2} dx\)

Step 8 ：\(\int \frac{5 x^{4}+7 x^{2}+x+2}{x\left(x^{2}+1\right)^{2}} d x = 5\int \frac{1}{x} dx + 3\int \frac{x - 1}{x^2 + 1} dx - 3\int \frac{x - \frac{1}{3}}{(x^2 + 1)^2} dx\)

Step 9 ：\(\int \frac{5 x^{4}+7 x^{2}+x+2}{x\left(x^{2}+1\right)^{2}} d x = 5\ln|x| + 3\ln(x^2 + 1) - 3\int \frac{x - \frac{1}{3}}{(x^2 + 1)^2} dx\)

Step 10 ：\(u = x^2 + 1, du = 2x dx\)

Step 11 ：\(-\frac{3}{2} \int \frac{u - \frac{1}{3}}{u^2} du = -\frac{3}{2} \int \frac{1}{u} du + \frac{1}{2} \int \frac{1}{u^2} du\)

Step 12 ：\(-\frac{3}{2} \int \frac{1}{u} du + \frac{1}{2} \int \frac{1}{u^2} du = -\frac{3}{2} \ln|u| + \frac{1}{2u} + C\)

Step 13 ：\(-\frac{3}{2} \ln|x^2 + 1| + \frac{1}{2(x^2 + 1)} + C = -\frac{3}{2} \ln(x^2 + 1) + \frac{1}{2(x^2 + 1)} + C\)

Step 14 ：\(\int \frac{5 x^{4}+7 x^{2}+x+2}{x\left(x^{2}+1\right)^{2}} d x = 5\ln|x| + 3\ln(x^2 + 1) - \left(-\frac{3}{2} \ln(x^2 + 1) + \frac{1}{2(x^2 + 1)} + C\right)\)

Step 15 ：\(\int \frac{5 x^{4}+7 x^{2}+x+2}{x\left(x^{2}+1\right)^{2}} d x = 5\ln|x| + \frac{9}{2} \ln(x^2 + 1) - \frac{1}{2(x^2 + 1)} + C\)

Step 16 ：\boxed{\int \frac{5 x^{4}+7 x^{2}+x+2}{x\left(x^{2}+1\right)^{2}} d x = 5\ln|x| + \frac{9}{2} \ln(x^2 + 1) - \frac{1}{2(x^2 + 1)} + C}\)