$\therefore {\rm{I}} = 2\int {\mathop x\limits_{II} } \mathop {{{\tan }^{ - 1}}}\limits_I xdx$
Applying Integration by parts
$\mathrm{I}=2\left[\tan ^{-1} x \int x d x-\int\left(\frac{d}{d x}\left(\tan ^{-1} x\right) \int x d x\right) d x\right]$
$\mathrm{I}=2\left[\frac{x^{2}}{2} \tan ^{-1} x-\int \frac{1}{1+x^{2}} \times \frac{x^{2}}{2} d x\right]+c$
${\rm{I}} = 2\left[ {\frac{{{x^2}}}{2}{{\tan }^{ - 1}}x - \frac{1}{2}\int {\frac{{{x^2} + 1 - 1}}{{{x^2} + 1}}} dx} \right] + c$
$I = 2\left[ {\frac{{{x^2}}}{2}{{\tan }^{ - 1}}x - \frac{1}{2}\int {\frac{{{x^2} + 1}}{{{x^2} + 1}}} dx + \frac{1}{2}\int {\frac{1}{{1 + {x^2}}}} dx} \right] + c$
$I = 2\left[ {\frac{{{x^2}}}{2}{{\tan }^{ - 1}}x - \frac{1}{2}\int {1.dx + \frac{1}{2}{{\tan }^{ - 1}}x} } \right] + c$
$1=2\left[\frac{x^{2}}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x\right]+c$
$I=x^{2} \tan ^{-1} x+\tan ^{-1} x-x+c$
or $\boxed{l = - x + ({x^2} + 1) t a n{ ^{ - 1}}x + c}$
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