Convolution Laplace transform 
Find the inverse Laplace transform of the giveb function by using the
  convolution theorem.
$$F(x) = \frac{s}{(s+1)(s^2+4)}$$

If I use partial fractions I get: $$\frac{s+4}{5(s^2+4)} - \frac{1}{5(x+1)}$$
which gives me Laplace inverses:
$$\frac{1}{5}(\cos2t + \sin2t) -\frac{1}{5} e^{-t}$$
But the answer is:
$$f(t) = \int^t_0 e^{-(t -\tau)}\cos(2\tau) d\tau$$
How did they get that?

Related techniques (I), (II). Using the fact about the Laplace transform $L$ that
$$ L(f*g)=L(f)L(g)=F(s)G(s)\implies  (f*g)(t)=L^{-1}(F(s)G(s)) .$$
In our case, given $ H(s)=\frac{1}{(s+1)}\frac{s}{(s^2+4)}$
$$F(s)=\frac{1}{s+1}\implies f(t)=e^{-t},\quad G(s)=\frac{s}{s^2+4}\implies g(t)=\cos(2t).$$
Now, you use the convolution as
$$ h(t) = \int_{0}^{t} e^{-(t-\tau)}\cos(2\tau) d\tau . $$  
