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I am asked to find the inverse Laplace transform of:

$$F(s)=\frac{2s^2+s+13}{(s-1)((s+1)^2+4)}$$

I did the partial fraction decomposition by hand and got:

$$F(s)=\frac{2}{s-1}-\frac{3}{(s+1)^2+4}$$

Mathematica's Apart command returned the same answer.

Apart[(2 s^2 + s + 13)/((s - 1) ((s + 1)^2 + 4))]

(* 2/(-1 + s) - 3/(5 + 2 s + s^2) *)

The answer can be written in the form:

$$F(s)=2\cdot\frac{1}{s-1}-\frac32\cdot\frac{2}{(s+1)^2+4}$$

I can then produce the inverse Laplace transform by hand:

$$f(t)=2e^t-\frac32e^{-t}\sin 2t$$

However, when I try to check my answer in Mathematica, I get the following response:

InverseLaplaceTransform[(2 s^2 + s + 13)/((s - 1) ((s + 1)^2 + 4)), s,
  t]

(* 2 E^t + 3/4 I E^((-1 - 2 I) t) (-1 + E^(4 I t)) *)

They are the same answer:

2 E^t + 3/4 I E^((-1 - 2 I) t) (-1 + E^(4 I t)) == 
  2 Exp[t] - 3/2*Exp[-t] Sin[2 t] // Simplify

(* True *)

But I am wondering how my hand calculated answer can be returned without the complex form.

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InverseLaplaceTransform[(2 s^2 + s + 13)/((s - 1) ((s + 1)^2 + 4)), s, t]

2*E^t + (3/4)IE^((-1 - 2*I)t) (-1 + E^(4*I*t))

% // ComplexExpand // Simplify

2*E^t - (3*Cos[t]*Sin[t])/E^t

% // TrigReduce

((1/2)*(4*E^(2*t) - 3*Sin[2*t]))/E^t

In "one" step

InverseLaplaceTransform[(2 s^2 + s + 13)/((s - 1) ((s + 1)^2 + 4)), s, t] // 
  ComplexExpand // TrigReduce

((1/2)*(4*E^(2*t) - 3*Sin[2*t]))/E^t

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  • $\begingroup$ Works! Thanks for the nice answer. $\endgroup$ – David Apr 30 '15 at 19:52

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