Given,

$$\int_1^e {{{{{\left( {\log _e^x} \right)}^{1/2}}} \over {x{{\left( {a - {{\left( {\log _e^x} \right)}^{3/2}}} \right)}^2}}} = 1} $$

Let $$\log _e^x = t$$

$$ \Rightarrow x = {e^t}$$

$$ \Rightarrow dx = {e^t}dt$$

When $$x = 1$$ then $$t = \log _e^1 = 0$$

and when $$x = e$$ then $$t = \log _e^e = 1$$

$$ \Rightarrow \int_0^1 {{{\sqrt t \,.\,{e^t}dt} \over {{e^t}{{(a - t\sqrt t )}^2}}} = 1} $$

$$ \Rightarrow \int_0^1 {{{\sqrt t \,dt} \over {{{(a - t\sqrt t )}^2}}} = 1} $$

Let $${t^{1/2}} = z$$

$$ \Rightarrow {1 \over {2\sqrt t }}dt = dz$$

$$ \Rightarrow dt = 2\sqrt t \,dz \Rightarrow dt = 2z\,dz$$

When $$t = 0$$, then $$z = 0$$

When $$t = 1$$, then $$z = {1^{1/2}} = 1$$

$$\therefore$$ $$\int_0^1 {{{z \times 2z\,dz} \over {{{(a - {z^3})}^2}}} = 1} $$

$$ \Rightarrow \int_0^1 {{{2{z^2}\,dz} \over {{{(a - {z^3})}^2}}} = 1} $$

Now let $$a - {z^3} = p$$

$$ \Rightarrow - 3{z^2}\,dz = dp$$

When $$z = 0$$ then $$p = a - 0 = a$$

When $$z = 1$$ then $$p = a - {1^3} = a - 1$$

$$\therefore$$ $$\int_a^{a - 1} {{{2\,.\,\left( { - {{dp} \over 3}} \right)} \over {{p^2}}} = 1} $$

$$ \Rightarrow - {2 \over 3}\int_a^{a - 1} {{{dp} \over {{p^2}}} = 1} $$

$$ \Rightarrow - {2 \over 3}\left[ { - {1 \over p}} \right]_a^{a - 1} = 1$$

$$ \Rightarrow - {2 \over 3}\left[ { - {1 \over {a - 1}} + {1 \over a}} \right] = 1$$

$$ \Rightarrow - {2 \over 3}\left[ {{{ - a + a - 1} \over {a(a - 1)}}} \right] = 1$$

$$ \Rightarrow {2 \over 3}\left( {{1 \over {a(a - 1)}}} \right) = 1$$

$$ \Rightarrow 3{a^2} - 3a = 2$$

$$ \Rightarrow 3{a^2} - 3a - 2 = 0$$

$$ \Rightarrow a = {{3\, \pm \,\sqrt {9 - 4(3)( - 2)} } \over {2\,.\,3}}$$

$$\therefore$$ $$a = {{3\, \pm \,\sqrt {33} } \over 6},\,{{3 - \sqrt {33} } \over 6}$$