If $$\int\limits_{{{\log }_e}2}^x {{{({e^x} - 1)}^{ - 1}}dx = {{\log }_e}{3 \over 2}} $$, then the value of x is

The value of $$\int\limits_0^5 {\max \{ {x^2},6x - 8\} \,dx} $$ is

Let f(x) be continuous periodic function with period T. Let $$I = \int\limits_a^{a + T} {f(x)\,dx} $$. Then

If $$b = \int\limits_0^1 {{{{e^t}} \over {t + 1}}dt} $$, then $$\int\limits_{a - 1}^a {{{{e^{ - t}}} \over {t - a - 1}}} $$ is