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# Integrate : $\int \limits _0^8 \tan ^{-1} x.dx$

$\begin {array} {1 1} (a)\;2 \pi-4 \log 2 \\ (b)\;2 \pi - \log 8 \\ (c)\;\pi-4 \log 2 \\ (d)\;2 \pi -3 \log 2 \end {array}$

x is the periodic function of period.
$\therefore \int \limits_0^8 \tan ^{-1} x dx$
$\qquad= 8 \bigg[ x \tan ^{-1} x -\large\frac{\log (1+x^2)}{2} \bigg]_0^1$
$\qquad = 8 \bigg[ \large\frac{\pi}{4} -\frac{ln2}{2}\bigg]$
$\qquad= 2 \pi -4 \log 2$
Hence a is the correct answer.