I=∫log1+1xx(1+x) dx=∫log1+1xx21x+1 dxPut log1+1x=t11+1x0+-1x2dx=dt==∫t-dt =-t22+C=-12log1+1x2+C =-12logx+1x2+C =-12logx+1-logx2+C =-12logx+12+logx2-2logx+1logx =-12logx+12-12logx2+logx+1logx+C

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$\int \frac{\log (1 + \frac{1}{x})}{x (1 + x)} d x =$

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$- \frac{1}{2} {[\log (x + 1)]}^{2} - \frac{1}{2} {[\log x]}^{2} + \log x . \log (x + 1) + C$

$\frac{1}{2} {[\log (x + 1)]}^{2} + \frac{1}{2} {[\log x]}^{2} - \log x . \log (x + 1) + C$

$- \frac{1}{2} {[\log (x + 1)]}^{2} - \frac{1}{2} {[\log x]}^{2} + \log [x (x + 1)] + C$

$\frac{1}{2} {[\log (x + 1)]}^{2} + \frac{1}{2} {[\log x]}^{2} - \log [x (x + 1)] + C$

answer is A.

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Detailed Solution

$I = \int \frac{\log (1 + \frac{1}{x})}{x (1 + x)} d x$

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

Put $\log (1 + \frac{1}{x}) = t$

$\frac{1}{1 + \frac{1}{x}} (0 + (\frac{- 1}{x^{2}})) d x = d t$

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