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Question: Find the value of the integral \[\int{\dfrac{{{e}^{x}}}{\sqrt{5-4{{e}^{x}}-{{e}^{2x}}}}dx}\] (a)...

Find the value of the integral
∫ex5−4ex−e2xdx\int{\dfrac{{{e}^{x}}}{\sqrt{5-4{{e}^{x}}-{{e}^{2x}}}}dx}
(a) cos⁡−1(ex+23)+c{{\cos }^{-1}}\left( \dfrac{{{e}^{x}}+2}{3} \right)+c
(b) cos⁡−1(ex−32)+c{{\cos }^{-1}}\left( \dfrac{{{e}^{x}}-3}{2} \right)+c
(c) sin⁡−1(ex+23)+c{{\sin }^{-1}}\left( \dfrac{{{e}^{x}}+2}{3} \right)+c
(d) sin⁡−1(ex−32)+c{{\sin }^{-1}}\left( \dfrac{{{e}^{x}}-3}{2} \right)+c

Explanation

Solution

Hint: First of all, take ex=t{{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, ∫dxa2−x2=sin⁡−1xa+c\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 ∫ex5−4ex−e2xdx\int{\dfrac{{{e}^{x}}}{\sqrt{5-4{{e}^{x}}-{{e}^{2x}}}}dx}. Let us consider the integral given in the question.
I=∫ex5−4ex−e2xdx.....(i)I=\int{\dfrac{{{e}^{x}}}{\sqrt{5-4{{e}^{x}}-{{e}^{2x}}}}dx}.....\left( i \right)
Let us take ex=t{{e}^{x}}=t. We know that ddxex=ex\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}. So, by differentiating both the sides, we get,
exdx=dt{{e}^{x}}dx=dt
Now, by substituting x in terms of t in equation (i), we get,
I=∫dt5−4t−t2I=\int{\dfrac{dt}{\sqrt{5-4t-{{t}^{2}}}}}
We can also write the above integral as,
I=∫dt−(t2+4t−5)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=∫dt−(t2+4t−5)+4−4I=\int{\dfrac{dt}{\sqrt{-\left( {{t}^{2}}+4t-5 \right)+4-4}}}
I=∫dt−(t2+4t+4)+5+4I=\int{\dfrac{dt}{\sqrt{-\left( {{t}^{2}}+4t+4 \right)+5+4}}}
I=∫dt9−(t2+4t+22)I=\int{\dfrac{dt}{\sqrt{9-\left( {{t}^{2}}+4t+{{2}^{2}} \right)}}}
We know that a2+b2+2ab=(a+b)2{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}. By using this, we get,
I=∫dt(3)2−(t+2)2I=\int{\dfrac{dt}{\sqrt{{{\left( 3 \right)}^{2}}-{{\left( t+2 \right)}^{2}}}}}
We know that ∫dxa2−x2=sin⁡−1xa+c\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}={{\sin }^{-1}}\dfrac{x}{a}+c}. By using this, we get,
I=sin⁡−1(t+2)3+cI={{\sin }^{-1}}\dfrac{\left( t+2 \right)}{3}+c
By separating t by ex{{e}^{x}}, we get,
I=sin⁡−1(ex+23)+cI={{\sin }^{-1}}\left( \dfrac{{{e}^{x}}+2}{3} \right)+c
Hence, option (c) is the right answer.

Note: In this question of integration containing ex{{e}^{x}}, it is always advisable to take ex{{e}^{x}} as t. Also, some students make this mistake of taking the integration of ∫dxx2−a2=1asin⁡−1xa+c\int{\dfrac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}=\dfrac{1}{a}{{\sin }^{-1}}\dfrac{x}{a}+c} while actually, it is sin⁡−1xa+c{{\sin }^{-1}}\dfrac{x}{a}+c. Also, students can cross-check their answer by differentiating sin⁡−1(ex+23)+c{{\sin }^{-1}}\left( \dfrac{{{e}^{x}}+2}{3} \right)+c and checking if it is equal to the expression given initially or not.