## Calculus 8th Edition

$$f^{\prime}(t)=t+\frac{1}{t^{3}}, \quad t\gt0 , \quad f(1)=6$$ The required function is $$f(t)=\frac{1}{2}t^{2}-\frac{1}{2}t^{-2}+6$$
$$f^{\prime}(t)=t+\frac{1}{t^{3}}, \quad t\gt0 , \quad f(1)=6$$ 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 $$f(t)=\frac{1}{2}t^{2}-\frac{1}{2}t^{-2}+6$$