Respuesta :
If you mean
[tex]\dfrac{6x+7}{(x+2)^2}[/tex]
the decomposition would be of the form
[tex]\dfrac{6x+7}{(x+2)^2} = \dfrac a{x+2} + \dfrac b{(x+2)^2}[/tex]
We have
[tex]6x+7 = 6(x+2) - 5[/tex]
which gives
[tex]\dfrac{6x+7}{(x+2)^2} = \dfrac{6(x+2)-5}{(x+2)^2} = \boxed{\dfrac6{x+2} - \dfrac5{(x+2)^2}}[/tex]
Or, if you prefer the standard approach, combining the partial fractions on the right side of the equation above would have given
[tex]\dfrac{6x+7}{(x+2)^2} = \dfrac{a(x+2)+b}{(x+2)^2}[/tex]
so that
[tex]6x+7 = a(x+2)+b = ax + 2a+b[/tex]
Solve for a and b :
[tex]\begin{cases}a=6\\2a+b=7\end{cases} \implies a=6,b=-5[/tex]