Math Help - LaPlace transforms

1. LaPlace transforms

Bear with me here, I took DIFF EQ 3 semesters ago and laplace transforms have come up again in a new class and I need to help
Solve the following using Laplace transforms:
$x''+x'+x=1$ where x $'(0)= 0$ and $x(0)=0$

I get the equation down to $X(s)=X(s^2+s+1)=\dfrac 1s$
then $X(s) = \dfrac 1{s(s^2+s+1)}$
next I went to partial fractions: $\dfrac As + \dfrac{Bs+C}{(s^2+s+1)}$
then finding a common denominator and setting it equal to 1 i get $1 = s^2(A+B) + s(A+C) + A$
Solving that I get $A=1,B=-1$ and $C=-1$

Plugging these back in I get $\dfrac 1s - \dfrac {s+1}{(s^2+s+1)}$
Now this is where I am stuck....I my table I cannot find a tranfer related to $\dfrac{s+1}{(s^2+s+1)}$ and I do not know how to manipulate it so I can use the tables.

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3. deleted

4. With all due respect to FernandoRevilla, I think the OP has the correct X(s). [EDIT]: FR has deleted his post. My advice to the OP'er is to complete the square in the denominator of the right-hand fraction, and also break up that fraction into the sum of two simpler fractions. The result will lead you, I think, to trig functions.

5. Originally Posted by tactical
in which step is this mistake?
The mistake was mine .

Express:

$s^2+s+1=\left(s+1/2\right)^2+3/4$

6. Originally Posted by Ackbeet
With all due respect to FernandoRevilla, I think the OP has the correct X(s). [EDIT]: FR has deleted his post.
Editing and answering sometimes have non empty intersecction.

Fernando Revilla

7. Originally Posted by FernandoRevilla
Editing and answering sometimes have non empty intersecction.
Aha.

8. Originally Posted by FernandoRevilla
The mistake was mine .

Express:

$s^2+s+1=\left(s+1/2\right)^2+3/4$
okay that make sense now. Thank you

9. You're certainly welcome for my contribution.