How to evaluate $\int\frac1{t^2}\exp(\int\frac1t\,\mathrm dt)\,\mathrm dt$? I want to evaluate the following integral:$$x(t)=\int\frac1{t^2}\exp\left(\int\frac1t\,\mathrm dt\right)\,\mathrm dt,$$
where $t$ is any value in $\mathbb{R}\backslash\{0\}$.
The next step is
$$x(t)=\int\frac1{t^2}\exp(\ln|t|)\,\mathrm dt=\int\frac{|t|}{t^2}\,\mathrm dt.$$
I am not sure how to continue from this step. The "correct solution" is $\ln(t)$, but no step is given. I do not see how to arrive at this solution. I am not given any additional information. Can I make some assumption?
 A: $\mathbb{R^*}=\mathbb{R}\setminus\{0\}$
$f:\mathbb{R^*}\to\mathbb{R}$ such that \begin{align} f(t)=\frac{\exp(\int\frac{1}{t}\ dt)}{t^2}=\frac{\exp(\ln(|t|)+c_0)}{t^2}=&\frac{c_1|t|}{t^2} \ & \mathbb{(c_0, c_1\in R).}\end{align} 
We will evaluate $\\[2ex]\int{f(t)}\,dt$.

Remember: $t^2>0$ for all $t\in\mathbb{R} \setminus \{0\}$. Similarly, we know that $|t|=\sqrt{t^2} > 0$ for all $t\in\mathbb{R} \setminus \{0\}$.
From those two characteristics, we will show that $0<\frac{|t|}{t^2}=\frac{1}{|t|}$ for all $t\in\mathbb{R} \setminus \{0\}$:
\begin{align} 
\frac{|t|}{t^2}=\frac{\sqrt{t^2}}{t^2}=\frac{\sqrt{t^2}\times\sqrt{t^2}}{t^2\times\sqrt{t^2}}=\frac{t^2}{t^2\times|t|}=\frac{1}{|t|} 
\end{align}

Note: $\frac{|t|}{t^2}=\frac{1}{t}$ only if we restrict our original domain to $\mathbb{R}_{>0}$, the set of positive real numbers.

We will use the fact that for any given $t$ in our domain, $\\[2ex]\frac{t}{|t|},\frac{|t|}{t}\in\{-1,1\}$ and $\frac{t}{|t|}\times\frac{|t|}{t}=1$:\begin{align} \int{f(t)}\,dt= & \int{\frac{c_1|t|}{t^2}}\,dt \\ & =c_1\int{\frac{|t|}{t^2}}\,dt \\ & =c_1\int{\frac{1}{|t|}}\,dt \\ & 
=c_1 \frac{t}{|t|} \int{\frac{1}{|t|}\frac{|t|}{t}}\,dt \\ & 
=c_1 \frac{t}{|t|}\int{\frac{1}{t}}\,dt \\ &
=c_1 \frac{t}{|t|} (\ln(|t|)+c_2)\end{align}
If we assume $c_0=0$, then $c_1=1$ and we arrive at $\frac{t}{|t|}\ln(|t|)+C$. 
