Answer
$$
f^{\prime}(t)=t+\frac{1}{t^{3}}, \quad t\gt0 , \quad f(1)=6
$$
The required function is
$$
f(t)=\frac{1}{2}t^{2}-\frac{1}{2}t^{-2}+6
$$
Work Step by Step
$$
f^{\prime}(t)=t+\frac{1}{t^{3}}, \quad t\gt0 , \quad f(1)=6
$$
can be written as the following:
$$
f^{\prime}(t)=t+\frac{1}{t^{3}}=t+t^{-3}
$$
The general anti-derivative of $
f^{\prime}(t)=t+\frac{1}{t^{3}}
$ is
$$
f(t)=\frac{1}{2}t^{2}-\frac{1}{2}t^{-2}+C
$$
To determine C we use the fact that $f(1)=6$:
$$
f(1)=\frac{1}{2}(1)^{2}-\frac{1}{2}(1)^{-2}+C =6
$$
$ \Rightarrow $
$$
0+C=6 \quad \Rightarrow \quad C=6,
$$
so
$$
f(t)=\frac{1}{2}t^{2}-\frac{1}{2}t^{-2}+6
$$
