I'm guessing your supposed to compute the value of the integral
[tex]I=\displaystyle\int_0^\infty\frac{e^{-t}\sin^2t}t\,\mathrm dt[/tex]
by way of Laplace transform.
The integral is equivalent to the transform of [tex]\frac{\sin^2t}t[/tex] when [tex]s=1[/tex], since
[tex]\mathscr L_s\left\{\dfrac{\sin^2t}t\right\}=\displaystyle\int_0^\infty\frac{\sin^2t}te^{-st}\,\mathrm dt[/tex]
Recall the frequency-domain integration property of the transform:
[tex]\mathscr L_s\left\{\dfrac{f(t)}t\right\}=\displaystyle\int_s^\infty F(\sigma)\,\mathrm d\sigma[/tex]
where [tex]F(s)[/tex] is the Laplace transform of [tex]f(t)[/tex].
Let [tex]f(t)=\sin^2t=\frac{1-\cos(2t)}2[/tex]; then
[tex]F(s)=\dfrac12\left(\dfrac1s-\dfrac s{s^2+4}\right)[/tex]
Next, it follows that
[tex]\mathscr L_s\left\{\dfrac{\sin^2t}t\right\}=\displaystyle\int_s^\infty F(\sigma)\,\mathrm d\sigma[/tex]
[tex]\displaystyle=\frac{\ln\sigma^2-\ln(\sigma^2+4)}4\bigg|_{\sigma=s}^{\sigma\to\infty}[/tex]
[tex]\displaystyle=\frac14\ln\left(1+\frac4{s^2}\right)[/tex]
Get the value of [tex]I[/tex] by substituting [tex]s=1[/tex]:
[tex]\displaystyle\int_0^\infty\frac{e^{-t}\sin^2t}t\,\mathrm dt=\boxed{\frac{\ln5}4}[/tex]
