Answer
Verify $\int\tan^2{x}dx=\tan{x}-x+C$
We need to take the derivative of the right side of our equation and verify that it equals the expression inside of our integral.
\begin{equation*}
\frac{d}{dx}\left(\tan{x}-x+C\right)=\frac{d}{dx}\left(\tan{x}\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(C\right)
\end{equation*}
\begin{equation*}
=\left(\sec^2{x}\right)-\left(1\right)+\left(0\right)
\end{equation*}
\begin{equation*}
=\tan^2{x}
\end{equation*}
as desired.
Work Step by Step
Verify $\int\tan^2{x}dx=\tan{x}-x+C$
We need to take the derivative of the right side of our equation and verify that it equals the expression inside of our integral.
\begin{equation*}
\frac{d}{dx}\left(\tan{x}-x+C\right)=\frac{d}{dx}\left(\tan{x}\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(C\right)
\end{equation*}
\begin{equation*}
=\left(\sec^2{x}\right)-\left(1\right)+\left(0\right)
\end{equation*}
\begin{equation*}
=\tan^2{x}
\end{equation*}
as desired.