∫log⁡1+1xx(1+x)dx=

∫log⁡1+1xx(1+x)dx Put log⁡1+1x=t, 11+1x−1x2dx=dt,−dxx(x+1)=dt=−t22+c=−12log⁡1+1x2+c=−12[log⁡(x+1)−log⁡x]2=−12[log⁡(x+1)]2−12[log⁡x]2+log⁡(x+1)log⁡x

∫log⁡1+1xx(1+x)dx=

∫log⁡1+1xx(1+x)dx Put log⁡1+1x=t, 11+1x−1x2dx=dt,−dxx(x+1)=dt=−t22+c=−12log⁡1+1x2+c=−12[log⁡(x+1)−log⁡x]2=−12[log⁡(x+1)]2−12[log⁡x]2+log⁡(x+1)log⁡x

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

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

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

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

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

answer is A.

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Detailed Solution

$\int \frac{\log (1 + \frac{1}{x})}{x (1 + x)} dx Put \log (1 + \frac{1}{x}) = t, \frac{1}{1 + \frac{1}{x}} (- \frac{1}{x^{2}}) dx = dt, - \frac{dx}{x (x + 1)} = dt$

$\begin{array}{l} = - \frac{t^{2}}{2} + c \\ = - \frac{1}{2} {(\log (1 + \frac{1}{x}))}^{2} + c \\ = - \frac{1}{2} [\log (x + 1) - \log x]^{2} \\ = - \frac{1}{2} [\log (x + 1)]^{2} - \frac{1}{2} [\log x]^{2} + \log (x + 1) \log x \end{array}$