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$\int\frac{\log(x+1)-\log x}{x(x+1)}dx$ is equal to: |
$-\frac{1}{2}[\log(\frac{x+1}{x})]^2+C$ $C-[\{\log(x+1)\}^2-(\log x)^2]$ $-\log[\log(\frac{x+1}{x})]+C$ $-\log(\frac{x+1}{x})+C$ |
$-\frac{1}{2}[\log(\frac{x+1}{x})]^2+C$ |
$\int\frac{\log(x+1)-\log x}{x(x+1)}dx$ $\int(\log(x+1)-\log x).\frac{1}{x(x+1)}dx=\frac{-1}{2}[\log(\frac{x+1}{x})]^2+C$ $[\frac{d}{dx}[(x-1)-\log x]=\frac{1}{x+1}-\frac{1}{x}=-\frac{1}{(x+1)x}]$ |
