# Find the Inverse Laplace Transform of the following function s+1/s^2+4

## Question:

Find the Inverse Laplace Transform of the following function

{eq}\frac{s+1}{{{s}^{2}}+4} {/eq}

## Inverse Laplace Transform

In mathematics, the Inverse Laplace of a function is exponentially restricted and continuous piecewise. Due to a large number of properties, it is useful to analyze the linear dynamical systems.

## Answer and Explanation:

Given Data

The given expression for the evaluation of inverse of Laplace is: {eq}\left( {\dfrac{{s + 1}}{{{s^2} + 4}}} \right) {/eq}.

The expression for the inverse of Laplace for the given expression is,

{eq}\begin{align*} f\left( t \right) &= {L^{ - 1}}\left( {\dfrac{{s + 1}}{{{s^2} + 4}}} \right)\\ f\left( t \right) &= {L^{ - 1}}\left( {\dfrac{s}{{{s^2} + 4}} + \dfrac{1}{{{s^2} + 4}}} \right)\\ f\left( t \right) &= {L^{ - 1}}\left( {\dfrac{s}{{{s^2} + {{\left( 2 \right)}^2}}}} \right) + {L^{ - 1}}\left( {\dfrac{1}{{{s^2} + {{\left( 2 \right)}^2}}}} \right)\\ f\left( t \right) &= \cos 2t + \dfrac{1}{2}{L^{ - 1}}\left( {\dfrac{2}{{{s^2} + {{\left( 2 \right)}^2}}}} \right) \end{align*} {/eq}

{eq}f\left( t \right) = \cos 2t + \dfrac{1}{2}\sin 2t {/eq}

Thus the inversion laplace of given expression is {eq}\cos 2t + \dfrac{1}{2}\sin 2t {/eq}.

#### Learn more about this topic: Differential Calculus: Definition & Applications

from Calculus: Help and Review

Chapter 13 / Lesson 6
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