What am I doing wrong in calculating the Laplace Inverse Transform of $\mathcal L^{-1} \left(\frac{3}{s}-\frac{3s}{s^2+1}\right)e^{-(s+1)x}$?

Question

Question: $$W=\left(\frac{3}{s}-\frac{3s}{s^2+1}\right)e^{-(s+1)x}$$

By taking the inverse laplace transform of $W(x,s)$ find $w(x,t)$?

My Attempt:

$$W=\left(\frac{3}{s}-\frac{3s}{s^2+1}\right)e^{-(s+1)x}$$

$$=\left(\color{orange}{\frac{3}{s}}-\color{red}{\frac{3s}{s^2+1}}\right)\cdot \color{purple}{e^{-sx}}\cdot \color{#06f}{e^{-x}}$$

Finding the Inverse Laplace Transform:

Since:

The Inverse Laplace Transform of:

$$\color{orange}{\frac{3}{s}=3\cdot 1}$$

and

$$\color{red}{\frac{3 \cdot s}{s^2+1}=3\cdot \cos t}$$

and

$$\frac{\color{purple}{e^{-sx}}}{s}=u(t-s) \implies \color{purple}{e^{-sx}=u(t-s) \cdot s}$$

and

$$\color{#06f}{e^{-x}=\frac{1}{s+1}}$$

We can use the 2nd Shifting Theorem, to get that the final answer is:

$$\bbox[10pt, border: blue solid 1pt]{\mathcal L^{-1}\left(\frac{3}{s}-\frac{3s}{s^2+1}\right)e^{-(s+1)x}=\frac{(\color{orange}{3}-\color{red}{3\cos(t)})\color{purple}{u(t-s)\cdot s}}{\color{#06f}{s+1}}}$$

But my Answer is wrong, what am I missing here?

Answer 1

Hint: $$\mathcal L^{-1}(1)=u(t)$$ Thus $$\mathcal L^{-1}(e^{-sx})=u(t-x)$$ And according to the 2nd Shift Theorem: $$\mathcal L^{-1}(F(s)*e^{-sx})=f(t-x)u(t-x)$$

Also $e^{-x}$is a constant. So rewriting the equation:

$$\mathcal L^{-1}\left(\frac{3}{s}-\frac{3s}{s^2+1}\right)e^{-(s+1)x}={-3\times(\color{red}{\cos(t-x)-1})\color{purple}{u(t-x)}}{}e^{-x}$$