[tex]\dfrac{dt}{dx} = x^2 + \frac{1}{64} \Rightarrow\ dt = \left(x^2 + \frac{1}{64}\right)dx \Rightarrow \\ \\ \displaystyle \int 1 dt = \int \left(x^2 + \frac{1}{64}\right)dx \Rightarrow \\ t = \dfrac{x^3}{3} + \frac{x}{64} + C[/tex]
C = 9 because all the x terms go away.
[tex]t = \dfrac{x^3}{3} + \dfrac{x}{64} + 9[/tex]