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Integrate the function
$\tan ^{-1} x$
$\tan ^{-1} x$
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Verified Answer
$\int \tan ^{-1} x d x=x \tan ^{-1} x-\frac{1}{2} \int \frac{2 x}{1+x^2} d x$
$=x \tan ^{-1} x-\frac{1}{2} \log \left|1+x^2\right|+C$
$=x \tan ^{-1} x-\frac{1}{2} \log \left|1+x^2\right|+C$
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