Answer
$f(t)=\displaystyle \frac{1}{2}\cdot t^{2}-\frac{1}{2}\cdot t^{-2}+6$
Work Step by Step
$\left[\begin{array}{ll}
\text{}f' & \text{particular antiderivative,}f\\
t^{1} & \frac{1}{2}\cdot t^{2}\\
& \\
+t^{-3} & -\frac{1}{2}\cdot t^{-2}
\end{array}\right]$
$f(t)=\displaystyle \frac{1}{2}\cdot t^{2}-\frac{1}{2}\cdot t^{-2}+C$
Given that $f(1)=6$, we find C:
$6=\displaystyle \frac{1}{2}\cdot(1^{2})-\frac{1}{2}\cdot(1^{-2})+C$
$C=6$
$f(t)=\displaystyle \frac{1}{2}\cdot t^{2}-\frac{1}{2}\cdot t^{-2}+6$
