CONCEPT:\(\smallint \frac{{{\rm{dx}}}}{{\rm{x}}} = \LN x + c\)CALCULATION:Let I = \(\smallint \frac{{{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}}}{{{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}}}{\rm{dx}}\)Let xe + ex = tDifferentiating both SIDES, we get⇒ (exe-1 + ex) dx = dt⇒ e (xe-1 + ex-1) dx = dt∴ (xe-1 + ex-1) dx = dt/eNow,\({\rm{I}} = \frac{1}{{\rm{e}}}\smallint \frac{{{\rm{dt}}}}{{\rm{t}}} = {\rm{\;}}\frac{1}{{\rm{e}}}\ln {\rm{t}} + {\rm{c}} = {\rm{\;}}\frac{1}{{\rm{e}}}{\rm{\;In\;}}\LEFT( {{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}} \right) + {\rm{c}}\)