Answer
$f(t)=\left\{\begin{array}{ll}
\dfrac{3^{t}}{\ln 3}-3\ln|t|+2-\dfrac{1}{3\ln 3}, & x\gt0\\ \\
\dfrac{3^{t}}{\ln 3}-3\ln|t|+1-\dfrac{3}{\ln 3}+3\ln 2, & x\lt0\\
\end{array}\right.$
Work Step by Step
$f'(t)=3^{t}-3\displaystyle \cdot\frac{1}{t}$
Using the antiderivatives table,
$f(t)=\displaystyle \frac{3^{t}}{\ln 3}-3\ln|t|+C$
The domain of f is $(-\infty,0)\cup (0,+\infty ).$
Given that
$f(1)=2\displaystyle \Rightarrow\quad 2=\frac{3^{-1}}{\ln 3}-3\ln|1|+C$
$C=2-\displaystyle \frac{1}{3\ln 3}$
(when $ x\gt0$ )
$f(-1)=1\displaystyle \Rightarrow\quad 1=\frac{3^{1}}{\ln 3}-3\ln|2|+C$
$C=1-\displaystyle \frac{3}{\ln 3}+3\ln 2$
(when $ x\lt0$ )
$f(t)=\left\{\begin{array}{ll}
\dfrac{3^{t}}{\ln 3}-3\ln|t|+2-\dfrac{1}{3\ln 3}, & x\gt0\\ \\
\dfrac{3^{t}}{\ln 3}-3\ln|t|+1-\dfrac{3}{\ln 3}+3\ln 2, & x\lt0\\
\end{array}\right.$
