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So Formula 6 gives
\begin{align*}
   \int_{0}^{1} \tan^{-1}/,dx &= x \tan^{-1}x\bigg]_{0}^{1}-\int_{0}^{1}\frac{x}{1+x^{2}}\,dx \\
   &=1\cdot\tan^{-1}0 - 0\cdot\tan^{-1}1 - \int_{0}^{1}\frac{x}{1+x{^2}}\,dx \\
   &=\frac{\pi}{4} - \int_{0}^{1}\frac{x}{1+x^{2}}\,dx. \\
\end{align*}
To evaluate this integral we use the substitution  $t=1+x^2$ (since $u$ has another meaning in this example).  Then  $dt=2x\,dx$, so $x\,dx=dt/2$.  When $x=0$, $t=1$.
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