The unilateral Laplace transform of f(t) is $${1 \over {{s^2} + s + 1}}.$$ Which one of the following is the unilateral Laplace transform of g(t) = t.f(t)?

The Laplace transform of $g(t) = t.f(t)$ is given by:

$$Lg(t): s = \int_0^\infty t.f(t)e^{-st}dt = \left {-{d \over ds}L[f(t): s} \right]_{s=0}$$

In this case, $Lf(t): s = {1 \over {{s^2} + s + 1}}$. Therefore,

$$Lg(t): s = \left[ {-{d \over ds} \left( {1 \over {{s^2} + s + 1}} \right)} \right]{s=0} = {-{d \over ds} \left( {s + 1 – {1 \over {s^2} + s + 1}} \right)} \right]{s=0}$$

Using the quotient rule, we can write the derivative of $Lf(t): s$ as:

$$L’f(t): s = {d \over ds} \left( {1 \over {{s^2} + s + 1}} \right) = {-{2s + 1} \over {{\left( {{s^2} + s + 1} \right)}^2}}$$

Therefore,

$$Lg(t): s = \left[ {-{d \over ds} \left( {s + 1 – {1 \over {s^2} + s + 1}} \right)} \right]_{s=0} = {-\left( {2s + 1} \right)} \over {{\left( {{s^2} + s + 1} \right)}^2}}$$

Therefore, the correct answer is $\boxed{{ – \left( {2s + 1} \right)} \over {{{\left( {{s^2} + s + 1} \right)}^2}}}$.