I need to find Laplace transform of this $f(t)$ $$f(t)=\frac{e^{at}-\cos bt}{t}$$
My approach: $$F(s)=\mathscr{L}\left(\frac{e^{at}}{t}\right)-\mathscr{L}\left(\text{Re}\left(\frac{e^{ibt}}{t}\right)\right)$$
But I don't know how to solve such cases. Please help.
That approach can't work, because $e^{at}/t$ does not have a Laplace transform, because $\int_0^1e^{at}/t\,dt=\infty$.
More or less as already suggested, if $F(s)=L[f]$ then $L[tf(t)]=-F'(s)$. This allows you to find $F'(s)$, which gives you $F(s)$ up to a "$+c"$. Then you can find the value of $c$ by considering what happens as $s\to+\infty$.
We can evaluate the Laplace Transform of $\frac{e^{at}-\cos(bt)}{t}$ directly without appealing to Feynman's Trick of differentiating under the integral.
METHODOLOGY $1$: GENERALIZED FRULANNI INTEGRAL
Note that we have
$$\begin{align} \int_0^\infty \frac{e^{at}-\cos(bt)}{t}\,e^{-st}\,dt&=\frac12\int_0^\infty \frac{e^{at}-e^{ibt}}{t}\,e^{-st}\,dt+\frac12\int_0^\infty \frac{e^{at}-e^{-ibt}}{t}\,e^{-st}\,dt\\\\ &=\frac12\int_0^\infty \frac{e^{-(s-a)t}-e^{-(s-ib)t}}{t}\,dt+\frac12\int_0^\infty \frac{e^{-(s-a)t}-e^{-(s+ib)t}}{t}\,dt\tag1 \end{align}$$
Applying the Generalized Frullani Integral (GFI), which I developed HERE, reveals
$$\begin{align} \int_0^\infty \frac{e^{at}-\cos(bt)}{t}\,e^{-st}\,dt&=\text{Re}\left(\log\left(\left|\frac{s-ib}{s-a}\right|\right)+i\arctan\left(b/(s-a)\right)\right)\\\\ &=\frac12\log\left(\frac{s^2+b^2}{(s-a)^2}\right) \end{align}$$
METHODOLOGY $2$: GENERALIZED FRULANNI INTEGRAL
Note that we have
$$\begin{align} \int_0^\infty \frac{e^{at}-\cos(bt)}{t}\,e^{-st}\,dt&=\int_0^\infty \frac{e^{at}-1}{t}\,e^{-st}\,dt+\int_0^\infty \frac{1-\cos(bt)}{t}\,e^{-st}\,dt\\\\ &=\int_0^\infty \frac{e^{-(s-a)t}-e^{-st}}{t}\,dt+\int_0^\infty \frac{1-\cos(bt)}{t}\,e^{-st}\,dt\tag2 \end{align}$$
The first integral on the right-hand side of $(2)$ is a Frullani integral and its value is
$$\int_0^\infty \frac{e^{-(s-a)t}-e^{-st}}{t}\,dt=\log\left(\frac{s}{s-a}\right)\tag3$$
for $s>a$.
The second integral can be written
$$\int_0^\infty \frac{1-\cos(bt)}{t}\,e^{-st}\,dt=\int_0^\infty \frac{1-\cos(t)}{t}e^{-(s/|b|)t}\,dt\tag4$$
In THIS ANSWER, I used only integration by parts and the identity (also proved at the end of THIS ANSWER) $\int_0^\infty \log(t)e^{-st}\,dt=\frac{-\gamma-\log(s)}{s}$, where $\gamma $ is the Euler-Mascheroni constant, to show that
$$\int_0^\infty \frac{1-\cos(t)}{t}e^{-st}\,dt=\frac12\log\left(\frac{s^2+1}{s^2}\right)\tag5$$
Using $(4)$ and $(5)$ reveals that
$$\int_0^\infty \frac{1-\cos(bt)}{t}\,e^{-st}\,dt=\frac12\log\left(\frac{s^2+b^2}{s^2}\right)\tag6$$
Using $(3)$ and $(6)$ in $(2)$ yields to coveted result
$$\int_0^\infty \frac{e^{at}-\cos(bt)}{t}\,e^{-st}\,dt=\frac12\log\left(\frac{s^2+b^2}{(s-a)^2}\right)$$
ALTERNITIVE DEVELOPMENT:
Here, we evaluate $(6)$ using the Generalized Frullani Integral (GFI) that I developed HERE.
We proceed by writing
$$\begin{align} \int_0^\infty \frac{1-\cos(bt)}{t}\,e^{-st}\,dt&=\int_0^\infty \frac{e^{-st}-\frac12\left(e^{-(s-ib)t}+e^{-(s+ib)t}\right)}{t}\,dt\\\\ &=\frac12 \int_0^\infty \frac{e^{-st}-e^{-(s-ibt)}}{t}\,dt+\frac12 \int_0^\infty \frac{e^{-st}-e^{-(s+ib)t}}{t}\,dt\\\\ &\overbrace{=}^{\text{GFI}}2\text{Re}\left(\frac12\log\left(\left|\frac{s-ib}{s}\right|\right)+i\arctan(b/s)\right)\\\\ &=\log\left(\sqrt{\frac{s^2+b^2}{s^2}}\right)\\\\ &=\frac12\log\left(\frac{s^2+b^2}{s^2}\right) \end{align}$$
which recovers $(6)$.
Just another way: $$\color{red}{\mathcal{L}_t\left[\frac{\exp (a t)-\cos (b t)}{t}\right](s)}=\\\mathcal{L}_t\left[\frac{\exp (a t)-\exp (i b t)}{t}\right](s)=\\\mathcal{L}_t\left[\int_0^{\infty } (\exp (a t)-\exp (i b t)) \exp (-k t) \, dk\right](s)=\\\int_0^{\infty } \mathcal{L}_t\left[e^{t (a-k)}-e^{t (i b-k)}\right](s) \, dk=\\\int_0^{\infty } \left(\frac{1}{-a+k+s}-\frac{1}{-i b+k+s}\right) \, dk=\\-\ln (-a+s)+\Re(\ln (-i b+s))=\\\color{red}{\frac{1}{2} \ln \left(b^2+s^2\right)-\ln (s-a)}$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}\bracks{\expo{at} - \cos\pars{bt} \over t}\expo{-st}\dd t}} = \Re\int_{0}^{\infty}{\expo{-\pars{s - a}t} - \expo{-\pars{s - \ic\verts{b}}t} \over t}\dd t \\[5mm] = &\ \Re\pars{-\int_{0}^{\infty}\ln\pars{t}\braces{% \expo{-\pars{s - a}t}\bracks{-\pars{s - a}} - \expo{-\pars{s - \ic\verts{b}}t}\bracks{-\pars{s - \ic\verts{b}}}}\dd t} \\[5mm] = &\ \Re\bracks{\int_{0}^{\infty}\ln\pars{t \over s - a}\expo{-t}\dd t - \int_{0}^{\infty}\ln\pars{t \over s - \ic\verts{b}}\expo{-t}\dd t}\label{1}\tag{1} \\[5mm] = &\ \Re\bracks{-\ln\pars{s - a} + \ln\pars{s - \ic\verts{b}}} = \bbx{{1 \over 2}\,\ln\pars{s^{2} + b^{2}} - \ln\pars{s - a}} \end{align}
The second integral in \eqref{1} comes from the integral $\ds{\int_{0}^{\color{red}{\large\pars{s - \ic\verts{b}}\infty}}\ln\pars{t \over s - \ic b}\expo{-t}\dd t}$ after a detour to the real axis. A remaining integral along an arc in the fourth quadrant of the complex plane vanishes out as the mentioned above arc radius $\ds{\to \infty}$.
It is possible to find the transform in the way you suggest if one uses distributions. If $t_+^{-1}$ is the distributional derivative of $(\ln t)_+$, then $$G(s) = \mathcal L[t_+^{-1}] = - \ln s - \gamma, \\ \mathcal L[t_+^{-1}(e^{a t} - \cos b t)] = G(s - a) - \frac {G(s + i b) + G(s - i b)} 2.$$
