Combine the fractions on the left with a common denominator:
[tex]\dfrac a{x+1} + \dfrac b{x-1} + \dfrac c{(x+1)^2} = \dfrac{a(x+1)(x-1) + b(x+1)^2 + c(x-1)}{(x-1)(x+1)^2}[/tex]
It follows that
[tex]3x^2+3x-2 = a(x+1)(x-1) + b(x+1)^2 + c(x-1)[/tex]
Expand the right side and collect like powers of x :
[tex]3x^2+3x-2 = a(x^2-1) + b(x^2+2x+1) + c(x-1)[/tex]
[tex]3x^2+3x-2 = (a+b)x^2 + (2b + c)x -a +b - c[/tex]
Then we have the system of equations
[tex]\begin{cases}a+b=3\\2b+c=3\\-a+b-c=-2\end{cases}[/tex]
or in matrix form,
[tex]\begin{bmatrix}1&1&0\\0&2&1\\-1&1&-1\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}3\\3\\-2\end{bmatrix}[/tex]
Compute the determinant of the coefficient matrix:
[tex]\det\begin{bmatrix}1&1&0\\0&2&1\\-1&1&-1\end{bmatrix} = -4[/tex]
Then the inverse of the coefficient matrix is equal 1/det times the adjugate of the coefficient matrix (a.k.a the transpose of the cofactor matrix):
[tex]\begin{bmatrix}1&1&0\\0&2&1\\-1&1&-1\end{bmatrix}^{-1} = \dfrac1{-4} \begin{bmatrix}-3 & -1 & 2 \\ 1 & -1 & -2 \\ 1 & -1 & 2\end{bmatrix}^\top = -\dfrac14 \begin{bmatrix}3&-1&-1\\1&1&1\\-2&2&-2\end{bmatrix}[/tex]
Multiply both sides of the equation by the inverse :
[tex]\begin{bmatrix}a\\b\\c\end{bmatrix} = -\dfrac14 \begin{bmatrix}3&-1&-1\\1&1&1\\-2&2&-2\end{bmatrix} \begin{bmatrix}3\\3\\-2\end{bmatrix} = \begin{bmatrix}2\\1\\1\end{bmatrix}[/tex]
So, we have a = 2 and b = c = 1, and the partial fraction decomposition is
[tex]\dfrac{3x^2+3x-2}{(x+1)^2(x-1)} = \dfrac 2{x+1} + \dfrac 1{x-1} + \dfrac 1{(x+1)^2}[/tex]
Answer:
Step-by-step explanation:
One can solve for a, b, c a little more directly than using a system of 3 equations.
If we multiply the rational expression by (x+1)², we get ...
(3x² +3x -2)/(x -1) = (x+1)²(a/(x+1) +b/(x-1)) +c
Evaluating this for x = -1 gives ...
(3(-1)² +3(-1) -2)/(-1 -1) = c
-2/-2 = 1 = c
Similarly, multiplying by (x -1) gives ...
(3x² +3x -2)/(x +1)² = (x -1)(a/(x +1) +c/(x +1)²) + b
Evaluating this for x = 1 gives ...
(3·1² +3·1 -2)/(1 +1)² = b
4/4 = 1 = b
Now, we need to find the value of 'a'. The identity will hold true for any value of x, so we can see what happens when we substitute x=0. We can use the values of 'b' and 'c' that we found above.
(3·0² +3·0 -2)/((0 +1)²(0 -1)) = a/(0 +1) +1/(0 -1) +1/(0 +1)²
-2/-1 = a -1 +1 ⇒ a = 2
_____
System of equations solution
When the terms of the right-side expansion are combined, the numerator of the result is ...
a(x +1)(x -1) +b(x +1)^2 +c(x -1) = (a+b)x² +(2b+c)x +(-a+b-c) ≡ 3x² +3x -2
Equating the coefficients gives the system of equations whose augmented matrix is:
[tex]\left[\begin{array}{ccc|c}1&1&0&3\\0&2&1&3\\-1&1&-1&-2\end{array}\right][/tex]
Transforming this to reduced row-echelon form using any of a variety of available tools gives ...
[tex]\left[\begin{array}{ccc|c}1&0&0&2\\0&1&0&1\\0&0&1&1\end{array}\right][/tex]
which tells you the solution is (A, B, C) = (2, 1, 1).
