Answer
$ f(t)=4\cdot\tan^{-1}t-\pi$
Work Step by Step
$\left[\begin{array}{cc}
\text{}f' & \text{particular antiderivative,}f\\
4\cdot\dfrac{1}{1+t^{2}} & 4\cdot\tan^{-1}t\\
&
\end{array}\right]$
$f(t)=4\cdot\tan^{-1}t+C$
Given that $f(1)=0$, we find C:
$0=4\cdot\tan^{-1}1+C$
$0=4\displaystyle \cdot\dfrac{\pi}{4}+C$
$ C=-\pi$
$ f(t)=4\cdot\tan^{-1}t-\pi$
