Question

What is the correct form of the partial fraction decomposition for 6x+7/(x+2)^2

Answer 1

If you mean

$\dfrac{6x+7}{(x+2)^2}$

the decomposition would be of the form

$\dfrac{6x+7}{(x+2)^2} = \dfrac a{x+2} + \dfrac b{(x+2)^2}$

We have

$6x+7 = 6(x+2) - 5$

which gives

$\dfrac{6x+7}{(x+2)^2} = \dfrac{6(x+2)-5}{(x+2)^2} = \boxed{\dfrac6{x+2} - \dfrac5{(x+2)^2}}$

Or, if you prefer the standard approach, combining the partial fractions on the right side of the equation above would have given

$\dfrac{6x+7}{(x+2)^2} = \dfrac{a(x+2)+b}{(x+2)^2}$

so that

$6x+7 = a(x+2)+b = ax + 2a+b$

Solve for a and b :

$\begin{cases}a=6\\2a+b=7\end{cases} \implies a=6,b=-5$