Answer
$f(t)=\displaystyle \frac{t^{4}}{12}-\ln|t|+\frac{8}{3}t+\ln 2-\frac{11}{3}$
Work Step by Step
$f''(t)=t^{2}+t^{-2}$
Using the antiderivatives table,
$f'(t)=\displaystyle \frac{t^{3}}{3}+\frac{t^{-1}}{-1}+C=\frac{1}{3}t^{3}-\frac{1}{t}+C$
Given that $f'(1)=2$ it follows that
$\displaystyle \frac{1}{3}(1)-1+C=2$
$C=\displaystyle \frac{8}{3}$
$f'(t)=\displaystyle \frac{1}{3}t^{3}-\frac{1}{t}+\frac{8}{3}$
Using the antiderivatives table,
$f(t)=\displaystyle \frac{1}{3}\cdot\frac{t^{4}}{4}-\ln|t|+\frac{8}{3}t+D$
$f(t)=\displaystyle \frac{t^{4}}{12}-\ln|t|+\frac{8}{3}t+D$
Given that $f(2)=3$ it follows that
$3=\displaystyle \frac{2^{4}}{12}-\ln|2|+\frac{8}{3}(2)+D$
$3=\displaystyle \frac{4}{3}-\ln 2+\frac{16}{3}+D$
$D=3-\displaystyle \frac{20}{3}+\ln 2=\ln 2-\frac{11}{3}$
$f(t)=\displaystyle \frac{t^{4}}{12}-\ln|t|+\frac{8}{3}t+\ln 2-\frac{11}{3}$
