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$\int\frac{\log_e x}{(1+ \log_e x)^2}dx$ is equal to |
$\frac{1}{(1+ \log_e x)^2}+C$, where C is constant of integration $\frac{x}{1+ \log_e x}+C$, where C is constant of integration $\frac{x}{(1+ \log_e x)^2}+C$, where C is constant of integration $\frac{1}{1+ \log_e x}+C$, where C is constant of integration |
$\frac{x}{1+ \log_e x}+C$, where C is constant of integration |
The correct answer is Option (2) → $\frac{x}{1+ \log_e x}+C$, where C is constant of integration $\int \frac{\log_e x}{(1+\log_e x)^2}\,dx$ Note: $\frac{d}{dx}\!\left(\frac{x}{1+\log_e x}\right)=\frac{(1+\log_e x)-1}{(1+\log_e x)^2}=\frac{\log_e x}{(1+\log_e x)^2}$ $\Rightarrow \int \frac{\log_e x}{(1+\log_e x)^2}\,dx=\frac{x}{1+\log_e x}+C$ $\frac{x}{1+\log_e x}+C$ |
