I suppose you mean the integral
[tex]\displaystyle \int \frac{x}{(x+1)^2 (x+2)^2} \, dx[/tex]
Break up the integrand into partial fractions:
[tex]\dfrac{x}{(x+1)^2 (x+2)^2} = \dfrac{a}{x+1} + \dfrac{b}{(x+1)^2} + \dfrac{c}{x+2} + \dfrac{d}{(x+2)^2}[/tex]
[tex]\dfrac{x}{(x+1)^2 (x+2)^2} \\ = \dfrac{a(x+1)(x+2)^2 + b(x+2)^2 + c(x+1)^2(x+2) + d(x+1)^2}{(x+1)^2 (x+2)^2}[/tex]
[tex]\dfrac{x}{(x+1)^2 (x+2)^2} \\ = \dfrac{(a+c)x^3+(5a+b+4c+d)x^2+(8a+4b+5c+2d)x + 4a+4b+2c+d}{(x+1)^2(x+2)^2}[/tex]
[tex]x = (a+c)x^3+(5a+b+4c+d)x^2+(8a+4b+5c+2d)x + 4a+4b+2c+d[/tex]
Solve for the coefficients:
[tex]\begin{cases}a+c=0 \\ 5a+b+4c+d=0 \\ 8a+4b+5c+2d=1 \\ 4a+4b+2c+d=0\end{cases} \implies a=3,b=-1,c=-3,d=-2[/tex]
So we have
[tex]\displaystyle \int \frac{x}{(x+1)^2 (x+2)^2} \, dx= \int \left(\frac3{x+1} - \frac1{(x+1)^2} - \frac3{x+2} - \frac2{(x+2)^2}\right) \, dx[/tex]
[tex]\displaystyle \int \frac{x}{(x+1)^2 (x+2)^2} \, dx= 3\ln|x+1| + \frac1{x+1} - 3\ln|x+2| + \frac2{x+2} + C[/tex]
[tex]\displaystyle \int \frac{x}{(x+1)^2 (x+2)^2} \, dx= \ln|x+1|^3 - \ln|x+2|^3 + \frac{3x+4}{(x+1)(x+2)} + C[/tex]
[tex]\displaystyle \int \frac{x}{(x+1)^2 (x+2)^2} \, dx= \boxed{\ln\left|\frac{x+1}{x+2}\right|^3 + \frac{3x+4}{(x+1)(x+2)} + C}[/tex]
