Question

The value of integral \[\int\limits_{0}^{\ln 5}{\dfrac{{{e}^{x}}\sqrt{{{e}^{x}}-1}}{{{e}^{x}}+3}dx}\] is
(a) \[4-\pi \]
(b) \[6-\pi \]
(c) \[5-\pi \]
(d) None of the above

Answer 1

Hint: We solve this problem by using the substitution method.
We assume the value inside the square root as a square of some other variable that is
\[{{e}^{x}}-1={{t}^{2}}\]
Then we differentiate the above equation to find the value of \[dx\] in terms of \[dt\]
Then we convert the given integral into a variable of \[t\] and also we change the limits of the integration according to the new variable.
We use the standard formula of integration that is
\[\int{\dfrac{1}{{{a}^{2}}+{{x}^{2}}}dx}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)\]

Complete step by step answer:
We are asked to find the value of integral \[\int\limits_{0}^{\ln 5}{\dfrac{{{e}^{x}}\sqrt{{{e}^{x}}-1}}{{{e}^{x}}+3}dx}\]
Let us assume that the given integral as
\[\Rightarrow I=\int\limits_{0}^{\ln 5}{\dfrac{{{e}^{x}}\sqrt{{{e}^{x}}-1}}{{{e}^{x}}+3}dx}........equation(i)\]
Now, let us use the substitution method of integration.
Let us assume that the value inside the root of given integral as some other variable that is
\[{{e}^{x}}-1={{t}^{2}}.......equation(ii)\]
Now, by differentiating the above equation on both sides with respect to \[x\] we get
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( {{e}^{x}}-1 \right)=\dfrac{d}{dx}\left( {{t}^{2}} \right) \\
 & \Rightarrow {{e}^{x}}=2t\dfrac{dt}{dx} \\
 & \Rightarrow {{e}^{x}}.dx=2t.dt \\
\end{align}\]
Now, let us change the limits of the given integral.
We are given that the lower limit as \[x=0\]
By substituting \[x=0\] in equation (ii) we get
\[\begin{align}
  & \Rightarrow {{e}^{0}}-1={{t}^{2}} \\
 & \Rightarrow {{t}^{2}}=1-1 \\
 & \Rightarrow t=0 \\
\end{align}\]
Similarly by substituting the upper limit that is \[x=\ln 5\] in equation (ii) we get
\[\Rightarrow {{e}^{\ln 5}}-1={{t}^{2}}\]
We know that the formula of exponents that is
\[{{e}^{\ln a}}=a\]
By using this condition in above equation we get
\[\begin{align}
  & \Rightarrow 5-1={{t}^{2}} \\
 & \Rightarrow t=2 \\
\end{align}\]
Now, we can say that the lower limit and upper limit of given integral with respect to \[t\] are 0 and 2
Now, by substituting the required values in terms of \[t\] in equation (i) we get
\[\begin{align}
  & \Rightarrow I=\int\limits_{0}^{2}{\dfrac{\sqrt{{{t}^{2}}}\left( 2tdt \right)}{\left( {{t}^{2}}+1 \right)+3}} \\
 & \Rightarrow I=\int\limits_{0}^{2}{\dfrac{2{{t}^{2}}}{{{t}^{2}}+4}dt} \\
 & \Rightarrow I=2\int\limits_{0}^{2}{\dfrac{{{t}^{2}}}{{{t}^{2}}+4}dt} \\
\end{align}\]
Now, let us try to remove the variable in the numerator.
So, by adding and subtracting 4 in the numerator we get
\[\Rightarrow I=2\int\limits_{0}^{2}{\dfrac{{{t}^{2}}+4-4}{{{t}^{2}}+4}dt}\]
Now, let us separate the terms so that we do not get variable in the numerator then we get
\[\begin{align}
  & \Rightarrow I=2\left[ \int\limits_{0}^{2}{\dfrac{{{t}^{2}}+4}{{{t}^{2}}+4}dt}+\int\limits_{0}^{2}{\dfrac{4}{{{t}^{2}}+4}dt} \right] \\
 & \Rightarrow I=2\int\limits_{0}^{2}{dt}+8\int\limits_{0}^{2}{\dfrac{1}{{{t}^{2}}+{{2}^{2}}}dt} \\
\end{align}\]
We know that the standard formulas of integration that are
\[\begin{align}
  & \int{dx}=x \\
 & \int{\dfrac{1}{{{a}^{2}}+{{x}^{2}}}dx}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right) \\
\end{align}\]
By using these two formulas in above equation we get
\[\Rightarrow I=2\left[ x \right]_{0}^{2}+8\left[ \dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]_{0}^{2}\]
Now, by applying the limits of integration in above equation we get
\[\begin{align}
  & \Rightarrow I=2\left( 2-0 \right)+8\left( \dfrac{1}{2}{{\tan }^{-1}}\dfrac{2}{2}-\dfrac{1}{2}{{\tan }^{-1}}\dfrac{0}{2} \right) \\
 & \Rightarrow I=4-4\left( \dfrac{\pi }{4} \right) \\
 & \Rightarrow I=4-\pi \\
\end{align}\]
Therefore we can conclude that the value of given integral is \[4-\pi \]
So, option (a) is correct answer.

Note:
Students may do mistakes after the assumption.
We assume that
\[{{e}^{x}}-1={{t}^{2}}\]
Here, we need to differentiate the above equation in order to get the value of \[dx\] in terms of \[dt\] then we get
\[\Rightarrow {{e}^{x}}.dx=2t.dt\]
But, students may do the mistake of not differentiating and directly assume that
\[\Rightarrow dx=dt\]
This will be wrong because when there is a change in the function then there will be a change in its differential part also.
Also, we need to change the limits to a new variable.
Original limits are given as 0 and \[\ln 5\] with respect to \[x\]
After the assumption, we need to change the limits to 0 and 2 with respect to \[t\]