I want to find the Laplace transform for this
$f(t)=2^t+\theta(t-1)\cos(t-1)$
The first part is easy:
\begin{equation} \int_0^\infty 2^te^{-st}dt=\int_0^\infty e^{t\ln2}e^{-st}dt=\int_0^\infty e^{t(\ln2-s)}dt=\bigg [\frac{1}{\ln2-s}e^{t(\ln2-s)}\bigg]_0^\infty=-\frac{1}{\ln2-s} \end{equation}
But the second, seems more tricky. Since the unit step function of this interval (t-1) is 0 for all $t<1$, we can readily form the given integral which is zero:
\begin{equation} \int_0^1 0\cos(t-1)e^{-st}dt=0 \end{equation}
but
Solve
\begin{equation} \int_1^\infty 1\cos(t-1)e^{-st}dt \end{equation}
\begin{equation} \int_1^\infty \cos(t-1)e^{-st}dt \end{equation}
But when trying integration by parts, we get:
\begin{equation} \int_1^\infty\cos(t-1)e^{-st}dt=-\frac{cos(t-1)}{s}e^{-st}-\frac{1}{s}\int_1^\infty\sin(t-1)e^{-st}dt \end{equation}
or we get
\begin{equation} \int_1^\infty\cos(t-1)e^{-st}dt=sin(t-1)e^{-st}+s\int_1^\infty\sin(t-1)e^{-st}dt \end{equation}
Doing an integration on the last term of the latter, gives:
\begin{equation} s\int_1^\infty\sin(t-1)e^{-st}dt=-\sin(t-1)e^{-st}+\int_1^\infty\cos(t-1)e^{-st}dt \end{equation}
Inserting this in the original integral:
\begin{equation} \int_1^\infty\cos(t-1)e^{-st}dt=sin(t-1)e^{-st}+\sin(t-1)e^{-st}-\int_1^\infty\cos(t-1)e^{-st}dt \end{equation}
which gives:
\begin{equation} 2\int_1^\infty\cos(t-1)e^{-st}dt=sin(t-1)e^{-st}+\sin(t-1)e^{-st} \end{equation}
finally
\begin{equation} \int_1^\infty\cos(t-1)e^{-st}dt=\frac{1}{2}\big(sin(t-1)e^{-st}+\sin(t-1)e^{-st}\big) \end{equation}
\begin{equation} \int_1^\infty\cos(t-1)e^{-st}dt=sin(t-1)e^{-st} \end{equation}
Thus:
The Laplace transform is
\begin{equation} \mathscr{L}\{2^t+\theta(t-1)\cos(t-1)\}=-\frac{1}{\ln2-s}+sin(t-1)e^{-st} \end{equation}
UPDATE
Using the second approach on that second term:
\begin{equation} \int_1^\infty\cos(t-1)e^{-st}dt=\frac{1}{2}\int_0^\infty(e^{iu}-e^{-iu})e^{-s(u+1)}du \end{equation}
\begin{equation} \frac{1}{2}\int_0^\infty(e^{iu}-e^{-iu})e^{-s(u+1)}du=\frac{1}{2}\int_0^\infty e^{iu-s(u+1)}-e^{-iu-s(u+1)}du \end{equation}
Take the former of the last two integrals
\begin{equation} \frac{1}{2}\int_0^\infty e^{iu-s(u+1)}du=\frac{1}{2}\bigg[\frac{e^{iu-s(u+1)}}{i-s}\bigg]_0^\infty \end{equation}
\begin{equation} \frac{1}{2}\bigg[\frac{e^{iu-s(u+1)}}{i-s}\bigg]_0^\infty=\lim_{u\rightarrow \infty}\frac{1}{2}\bigg[\frac{e^{iu-s(u+1)}}{i-s}\bigg]-\lim_{u\rightarrow 0}\frac{1}{2}\bigg[\frac{e^{iu-s(u+1)}}{i-s}\bigg] \end{equation}
\begin{equation} =-\frac{1}{2}\frac{e^{-s}}{i-s} \end{equation}
Take the latter of the two integrals,
\begin{equation} \frac{1}{2}\int_0^\infty e^{-iu-s(u+1)}du=\frac{1}{2}\bigg[\frac{e^{-iu-s(u+1)}}{-i-s}\bigg]_0^\infty \end{equation}
\begin{equation} \frac{1}{2}\bigg[\frac{e^{-iu-s(u+1)}}{-i-s}\bigg]_0^\infty=\frac{1}{2}\lim_{u\rightarrow \infty}\bigg[\frac{e^{-iu-s(u+1)}}{-i-s}\bigg]-\frac{1}{2}\lim_{u\rightarrow 0}\bigg[\frac{e^{-iu-s(u+1)}}{-i-s}\bigg] \end{equation}
\begin{equation} =-\frac{1}{2}\frac{e^{-s}}{-i-s} \end{equation}
\begin{equation} \mathscr{L}\{2^t+\theta(t-1)\cos(t-1)\}=-\frac{1}{\ln2-s}-\frac{e^{-s}}{2}\bigg(\frac{1}{i-s}+\frac{1}{-i-s}\bigg) \end{equation}
but this differs from the correct answer found by integration by parts
Sorry, I cannot comment in the comment section. What if you write $cos(t-1)=cos(t)cos(1)+sin(t)sin(1)$ and then use the polar form of the trigonometric functions? After distributions, you will get everything in terms of constants, exponentials, and products of the form $e^{at}t$, for suitable $a$ and for all of them you should have explicit formulas (remember that the exponential just moves the argument of the Laplace transform.
Usually in this kind of problems is not necessary to solve the integral, but rather play a little bit with the properties before getting something that can be computed using the standard Laplace table.
I hope this helps!