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

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

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

Step 3 ：$5x^4+7x^2+x+2=A{x}^4+2A{x}^2+A+B{x}^3+Bx+C{x}^2+C+D{x}^3+Dx+E{x}^2+E$

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

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

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

Step 10 ：$u=x^2+1,du=2xdx$

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 ：$\int \frac{5x^4+7x^2+x+2}{x{(x^2+1)}^2}dx=5\ln |x|+3\ln (x^2+1)-(-\frac{3}{2}\ln (x^2+1)+\frac{1}{2(x^2+1)}+C)$

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

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}\)