Let $$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ be a differentiable function for all x$$\in$$R. Then f(x) equals :

For x > 0, if $$f(x) = \int\limits_1^x {{{{{\log }_e}t} \over {(1 + t)}}dt} $$, then $$f(e) + f\left( {{1 \over e}} \right)$$ is equal to :

The value of $$\int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$$ is :

The value of $$\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}dx} } $$, where [ x ] is the greatest integer $$ \le $$ x, is :