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Integrate the function
$x \tan ^{-1} x$
$x \tan ^{-1} x$
Solution:
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Verified Answer
$\int x \tan ^{-1} x d x=\tan ^{-1} x\left(\frac{x^2}{2}\right)-\int \frac{1}{1+x^2} \cdot \frac{x^2}{2} d x$
$=\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int\left(1-\frac{1}{1+x^2}\right) d x$
$=\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} x+\frac{1}{2} \tan ^{-1} x+C$
$=\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int\left(1-\frac{1}{1+x^2}\right) d x$
$=\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} x+\frac{1}{2} \tan ^{-1} x+C$
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