Finding the Inverse Laplace Transform of $-\frac{1}{s(s+1)}+\frac{(s+1)(s+2)}{s\left((s+1)^2+1\right)}$ 
I am trying to find the inverse Laplace transform of $$F(s)=-\frac{1}{s(s+1)}+\frac{(s+1)(s+2)}{s\left((s+1)^2+1\right)}.$$

I proceeded as follows:
\begin{align}
F(s)&=-\frac{1}{s(s+1)}+\frac{(s+1)(s+2)}{s\left((s+1)^2+1\right)} \\
&=-\frac{1}{s}+\frac{1}{s+1}+\frac{s+1}{(s+1)^2+1}+\frac{4}{s\left((s+1)^2+1\right)} \\
f(t)&=-1+e^{-t}+e^{-t}\cos(t)+\mathcal{L}^{-1}\left(\frac{4}{s}\times\frac{1}{(s+1)^2+1}\right) \\
&=-1+e^{-t}+e^{-t}\cos(t)+4\int_0^{\infty}e^{-u}\sin(u) \ du \ \ \ \ \ \text{(convolution theorem)} \\
&=1+e^{-t}(1+\cos(t))
\end{align}
But the answer is apparently $f(t)=e^{-t}(1+\sin(t)).$ I have looked over my worked and agree with my solution. 
edit:
\begin{align}
\frac{(s+1)(s+2)}{s\left((s+1)^2+1\right)}&=\frac{s^2+3s+2}{s\left((s+1)^2+1\right)} \\
&=\frac{s^2+3s}{s\left((s+1)^2+1\right)}+\frac{2}{s\left((s+1)^2+1\right)} \\
&=\frac{s(s+3)}{s\left((s+1)^2+1\right)}+\frac{2}{s\left((s+1)^2+1\right)} \\
&=\frac{s+3}{(s+1)^2+1}+\frac{2}{s\left((s+1)^2+1\right)} \\
&=\frac{s+1}{(s+1)^2+1}+\frac{4}{s\left((s+1)^2+1\right)}
\end{align}

Your partial fractions are not correct for the second term:
\begin{equation}
\frac{(s+1)(s+2)}{s((s+1)^2+1)}=\frac{1}{(s+1)^2+1}+\frac{1}{s}.
\end{equation}
Then,
\begin{equation}
F(s) = \frac{1}{(s+1)}+\frac{1}{(s+1)^2+1}.
\end{equation}
Thus, you get the desired result!
