## Calculus: Early Transcendentals 8th Edition

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