The problem asks us to resolve the rational function $\frac{4x-3}{(x+1)^2}$ into partial fractions.

1. Problem Description

The problem asks us to resolve the rational function $\frac{4x-3}{(x+1)^2}$ into partial fractions.

2. Solution Steps

Since the denominator is $(x+1)^2$, which is a repeated linear factor, we can express the given fraction as:

$\frac{4x-3}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}$

To find the values of A and B, we multiply both sides of the equation by $(x+1)^2$:

$4x-3 = A(x+1) + B$

$4x-3 = Ax + A + B$

Now we can equate the coefficients of the terms with the same power of $x$.

Equating the coefficients of $x$, we have:

$4 = A$

Equating the constant terms, we have:

$-3 = A + B$

Since we know $A=4$, we can substitute this value into the second equation:

$-3 = 4 + B$

$B = -3 - 4$

$B = -7$

Therefore, $A = 4$ and $B = -7$. Substituting these values into the partial fraction decomposition, we get:

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

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

3. Final Answer

The partial fraction decomposition is: $\frac{4}{x+1} - \frac{7}{(x+1)^2}$.