If ∫x(1+x)log(1+x2)dx=f(x)log(1+x2)-23tan-1x-2x39-x22+2x3+C then fx=

∫x(1+x)log(1+x2)dx ∫log1+x2.x+x2dx =log1+x2x22+x33-∫11+x22xx22+x33dx=log1+x2x22+x33-∫x31+x2dx-23∫x4x2+1 =log1+x2x22+x33-∫x-xx2+1dx -23∫x2-1+1x2+1dx =log1+x2x22+x33-x22+12logx2+1 -23∫x2-1x2+1x2+1+1x2+1dx =log1+x2x22+x33-x22+12logx2+1 -23∫x2-1+1x2+1dx =log1+x2x22+x33+12-x22 -23x33-x+tan-1x ⇒fx=x22+x33+12

If ∫x(1+x)log(1+x2)dx=f(x)log(1+x2)-23tan-1x-2x39-x22+2x3+C then fx=

∫x(1+x)log(1+x2)dx ∫log1+x2.x+x2dx =log1+x2x22+x33-∫11+x22xx22+x33dx=log1+x2x22+x33-∫x31+x2dx-23∫x4x2+1 =log1+x2x22+x33-∫x-xx2+1dx -23∫x2-1+1x2+1dx =log1+x2x22+x33-x22+12logx2+1 -23∫x2-1x2+1x2+1+1x2+1dx =log1+x2x22+x33-x22+12logx2+1 -23∫x2-1+1x2+1dx =log1+x2x22+x33+12-x22 -23x33-x+tan-1x ⇒fx=x22+x33+12

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If $\int x (1 + x) \log (1 + x^{2}) d x = f (x) \log (1 + x^{2})$ $- \frac{2}{3} \tan^{- 1} x - \frac{2 x^{3}}{9} - \frac{x^{2}}{2} + \frac{2 x}{3} + C$ then $f (x) =$

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$\frac{x^{2}}{2} + \frac{x^{4}}{3}$

$\frac{x^{2}}{2} + \frac{x^{2}}{3} - \frac{1}{3}$

$\frac{x^{2}}{2} + \frac{x^{2}}{3} + \frac{1}{2}$

$\frac{x^{2}}{2} + \frac{x^{2}}{3} - \frac{2}{3}$

answer is C.

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Detailed Solution

$\int x (1 + x) \log (1 + x^{2}) d x \int \log (1 + x^{2}) . (x + x^{2}) d x = \log (1 + x^{2}) (\frac{x^{2}}{2} + \frac{x^{3}}{3}) - \int \frac{1}{1 + x^{2}} (2 x) (\frac{x^{2}}{2} + \frac{x^{3}}{3}) d x$

$= \log (1 + x^{2}) (\frac{x^{2}}{2} + \frac{x^{3}}{3}) - \int \frac{x^{3}}{1 + x^{2}} d x - \frac{2}{3} \int \frac{x^{4}}{x^{2} + 1} = \log (1 + x^{2}) (\frac{x^{2}}{2} + \frac{x^{3}}{3}) - \int [x - \frac{x}{x^{2} + 1}] d x - \frac{2}{3} \int \frac{(x^{2} - 1) + 1}{x^{2} + 1} d x$

$= \log (1 + x^{2}) (\frac{x^{2}}{2} + \frac{x^{3}}{3}) - \frac{x^{2}}{2} + \frac{1}{2} \log (x^{2} + 1) \frac{- 2}{3} \int [\frac{(x^{2} - 1) (x^{2} + 1)}{x^{2} + 1} + \frac{1}{x^{2} + 1}] d x$

$= \log (1 + x^{2}) (\frac{x^{2}}{2} + \frac{x^{3}}{3}) - \frac{x^{2}}{2} + \frac{1}{2} \log (x^{2} + 1) - \frac{2}{3} \int (x^{2} - 1 + \frac{1}{x^{2} + 1}) d x = \log (1 + x^{2}) (\frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{1}{2}) - \frac{x^{2}}{2} - \frac{2}{3} (\frac{x^{3}}{3} - x + \tan^{- 1} x) \Rightarrow f (x) = \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{1}{2}$