Find the value of the integral
\[\int{\dfrac{{{e}^{x}}}{\sqrt{5-4{{e}^{x}}-{{e}^{2x}}}}dx}\]
(a) \[{{\cos }^{-1}}\left( \dfrac{{{e}^{x}}+2}{3} \right)+c\]
(b) \[{{\cos }^{-1}}\left( \dfrac{{{e}^{x}}-3}{2} \right)+c\]
(c) \[{{\sin }^{-1}}\left( \dfrac{{{e}^{x}}+2}{3} \right)+c\]
(d) \[{{\sin }^{-1}}\left( \dfrac{{{e}^{x}}-3}{2} \right)+c\]

Hint: First of all, take \[{{e}^{x}}=t\] and write the given integral in terms of t. Now, add and subtract 4 from the denominator to make a perfect square in the denominator. Now use, \[\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}={{\sin }^{-1}}\dfrac{x}{a}+c}\] to get the required answer.

Complete step-by-step solution -
In this question, we have to find the value of the integral \[\int{\dfrac{{{e}^{x}}}{\sqrt{5-4{{e}^{x}}-{{e}^{2x}}}}dx}\]. Let us consider the integral given in the question.
\[I=\int{\dfrac{{{e}^{x}}}{\sqrt{5-4{{e}^{x}}-{{e}^{2x}}}}dx}.....\left( i \right)\]
Let us take \[{{e}^{x}}=t\]. We know that \[\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}\]. So, by differentiating both the sides, we get,
\[{{e}^{x}}dx=dt\]
Now, by substituting x in terms of t in equation (i), we get,
\[I=\int{\dfrac{dt}{\sqrt{5-4t-{{t}^{2}}}}}\]
We can also write the above integral as,
\[I=\int{\dfrac{dt}{\sqrt{-\left( {{t}^{2}}+4t-5 \right)}}}\]
By adding and subtracting 4 from the denominator of the above equation, we get,
\[I=\int{\dfrac{dt}{\sqrt{-\left( {{t}^{2}}+4t-5 \right)+4-4}}}\]
\[I=\int{\dfrac{dt}{\sqrt{-\left( {{t}^{2}}+4t+4 \right)+5+4}}}\]
\[I=\int{\dfrac{dt}{\sqrt{9-\left( {{t}^{2}}+4t+{{2}^{2}} \right)}}}\]
We know that \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\]. By using this, we get,
\[I=\int{\dfrac{dt}{\sqrt{{{\left( 3 \right)}^{2}}-{{\left( t+2 \right)}^{2}}}}}\]
We know that \[\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}={{\sin }^{-1}}\dfrac{x}{a}+c}\]. By using this, we get,
\[I={{\sin }^{-1}}\dfrac{\left( t+2 \right)}{3}+c\]
By separating t by \[{{e}^{x}}\], we get,
\[I={{\sin }^{-1}}\left( \dfrac{{{e}^{x}}+2}{3} \right)+c\]
Hence, option (c) is the right answer.

Note: In this question of integration containing \[{{e}^{x}}\], it is always advisable to take \[{{e}^{x}}\] as t. Also, some students make this mistake of taking the integration of \[\int{\dfrac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}=\dfrac{1}{a}{{\sin }^{-1}}\dfrac{x}{a}+c}\] while actually, it is \[{{\sin }^{-1}}\dfrac{x}{a}+c\]. Also, students can cross-check their answer by differentiating \[{{\sin }^{-1}}\left( \dfrac{{{e}^{x}}+2}{3} \right)+c\] and checking if it is equal to the expression given initially or not.