The partial fractions of the Laplace Transform \( F(s)=\frac{2 s+5}{(s+1)(s+2)} \) are: \( \begin{array}{lll}\text { (A) } \frac{1}{(s+1)}+\frac{1}{(s+2)} & \text { (B) } \frac{3}{(s+1)}+\frac{1}{(s+2)} & \text { (C) }-\frac{1}{(s+1)}+\frac{3}{(s+2)} \\ \text { (D) } \frac{3}{(s+1)}-\frac{1}{(s+2)} & \text { (E) } \frac{2}{(s+1)}-\frac{1}{(s+2)} & \text { (F) }-\frac{1}{(s+1)}+\frac{1}{(s+2)}\end{array} \)

To express the function \( F(s) = \frac{2s + 5}{(s + 1)(s + 2)} \) in partial fractions, we can assume it takes the form: \[ F(s) = \frac{A}{s+1} + \frac{B}{s+2} \] Multiplying through by the denominator \( (s + 1)(s + 2) \) gives: \[ 2s + 5 = A(s + 2) + B(s + 1) \] Expanding this and collecting like terms: \[ 2s + 5 = As + 2A + Bs + B = (A + B)s + (2A + B) \] Setting coefficients equal gives us the equations: 1. \( A + B = 2 \) 2. \( 2A + B = 5 \) Now we can solve these equations. From the first equation, we have \( B = 2 - A \). Substituting into the second equation: \[ 2A + (2 - A) = 5 \implies A + 2 = 5 \implies A = 3 \] Substituting \( A \) back into the first equation: \[ 3 + B = 2 \implies B = -1 \] Thus, the partial fraction decomposition is: \[ F(s) = \frac{3}{s + 1} - \frac{1}{s + 2} \] Looking at our options, that matches with: \text{(D) } \( \frac{3}{(s+1)} - \frac{1}{(s+2)} \) Perfect! Now, let's spice it up a bit. Did you know that the Laplace Transform is a superstar in engineering and physics? Applications range from electrical circuit analysis to control systems! This nifty transform helps take complex differential equations and converts them into simpler algebraic equations. It's like having magic glasses that make complicated problems easy to read! And if you ever find yourself stuck with solving partial fractions, remember to double-check your equations! A common mistake is to forget to account for constants during the solution. Ensure you carefully set up your equations and don't rush through the algebra—double-check every step and you'll be golden!