The value of \(\smallint \frac{{{e^x}}}{{{e^{2x}} - 4}}dx\)  will be ______, where C is an arbitrary constant.

Concept:

From the Standard integral:

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHRiI8daWcaaWdaeaapeGaamizaiaadIhaa8aabaWdbiaadIha % paWaaWbaaSqabeaapeGaaGOmaaaakiabgkHiTiaadggapaWaaWbaaS % qabeaapeGaaGOmaaaaaaGccqGH9aqpdaWcaaWdaeaapeGaaGymaaWd % aeaapeGaaGOmaiaadggaaaGaamiBaiaad+gacaWGNbWaaqWaa8aaba % Wdbmaalaaapaqaa8qacaWG4bGaeyOeI0IaamyyaaWdaeaapeGaamiE % aiabgUcaRiaadggaaaaacaGLhWUaayjcSdaaaa!4EB0! \smallint \frac{{dx}}{{{x^2} - {a^2}}} = \frac{1}{{2a}}log\left| {\frac{{x - a}}{{x + a}}} \right|\)\(\smallint \frac{{{dx}}}{{{x^2} - a^2}}=\frac{1}{2a}log \ |\frac{x-a}{x+a}|+C, \ x>a\)

Calculation:

\(\smallint \frac{{{e^x}}}{{{e^{2x}} - 4}}dx\)

\(\smallint \frac{{{e^x}}}{{({e^{x})^2} - (2)^2}}dx\)

let t = e^x

dt = e^x dx

\(\smallint \frac{{{dt}}}{{({t)^2} - (2)^2}}\)

From the standard integral:

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHRiI8daWcaaWdaeaapeGaamizaiaadIhaa8aabaWdbiaadIha % paWaaWbaaSqabeaapeGaaGOmaaaakiabgkHiTiaadggapaWaaWbaaS % qabeaapeGaaGOmaaaaaaGccqGH9aqpdaWcaaWdaeaapeGaaGymaaWd % aeaapeGaaGOmaiaadggaaaGaamiBaiaad+gacaWGNbWaaqWaa8aaba % Wdbmaalaaapaqaa8qacaWG4bGaeyOeI0IaamyyaaWdaeaapeGaamiE % aiabgUcaRiaadggaaaaacaGLhWUaayjcSdaaaa!4EB0! \smallint \frac{{dx}}{{{x^2} - {a^2}}} = \frac{1}{{2a}}log\left| {\frac{{x - a}}{{x + a}}} \right|\)\(\smallint \frac{{{dt}}}{{({t)^2} - (2)^2}}=\frac{1}{4}log \ |\frac{t-a}{t+a}|+C\)

Put t = ex in the above equation, we get:

\(\smallint \frac{{{e^x}}}{{{e^{2x}} - 4}}dx=\frac{1}{4}\log \left| {\frac{{{e^x} - 2}}{{{e^x} + 2}}} \right| + C\)

Note:

Some important formulas of integration are:

\(\smallint \frac{{{dx}}}{{{a^2} - x^2}}=\frac{1}{2a}log \ |\frac{a+x}{a-x}|+C, \ x<a\)

\(\smallint \frac{{{dx}}}{{{\sqrt{{a^2} - x^2}}}}=sin^{-1}(\frac{x}{a})+C\)

\(\smallint \frac{{{dx}}}{{{\sqrt{{x^2} + a^2}}}}=log (\ x + \sqrt{a^2+x^2})+C\)