If ∫x2Tan−1⁡x1+x2dx=xTan−1⁡x−12log⁡1+x2+f(x)+c then f(x)=

I=∫x2 tan−1⁡x1+x2dx Put tan-1x=t⇒11+x2dx=dt; I=∫tan2(t).t dt = ∫t.(sec2t-1)dt = ∫t.sec2t dt-t22+c = (t tant)-∫1.tant dt - t22+c = t tant - logsect -t22+c = (tan-1x)x-log(1+x2)-(tan-1x)22+c ∴ f(x) = -(tan-1x)22

If ∫x2Tan−1⁡x1+x2dx=xTan−1⁡x−12log⁡1+x2+f(x)+c then f(x)=

I=∫x2 tan−1⁡x1+x2dx Put tan-1x=t⇒11+x2dx=dt; I=∫tan2(t).t dt = ∫t.(sec2t-1)dt = ∫t.sec2t dt-t22+c = (t tant)-∫1.tant dt - t22+c = t tant - logsect -t22+c = (tan-1x)x-log(1+x2)-(tan-1x)22+c ∴ f(x) = -(tan-1x)22

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$If \int \frac{x^{2} {Tan}^{- 1} x}{1 + x^{2}} d x = x {Tan}^{- 1} x - \frac{1}{2} \log (1 + x^{2}) + f (x) + c then f (x) =$

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$- \frac{1}{2} {({Tan}^{- 1} x)}^{2}$

$- \frac{{Tan}^{- 1} x}{2}$

$\frac{{Tan}^{- 1} x}{2}$

$\frac{{({Tan}^{- 1} x)}^{2}}{2}$

answer is B.

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Detailed Solution

$I = \int \frac{x^{2} \tan^{- 1} x}{1 + x^{2}} d x$
$Put \tan^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} dx = dt; I = \int \tan^{2} (t) . t dt = \int t . (\sec^{2} t - 1) dt = \int t . \sec^{2} t dt - \frac{t^{2}}{2} + c = (t tant) - \int 1 . tant dt - \frac{t^{2}}{2} + c = t tant - logsect - \frac{t^{2}}{2} + c = (\tan^{- 1} x) x - \log (\sqrt{1 + x^{2}}) - \frac{{(\tan^{- 1} x)}^{2}}{2} + c ∴ f (x) = \frac{- {(\tan^{- 1} x)}^{2}}{2}$