Question

If $\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx = g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$ where $c$ is a constant of integration, then find the value of $g\left( 0 \right)$.
A. 2
B. e
C. 1
D. ${e^2}$

Answer 1

Hint: In the given question, first we will solve the integration. To solve the integration we need to divide the integration into two integrations such that the derivative of the exponent will be the coefficient of the integration. By using the substitution method we will solve it and compare both sides of the equation to get the value of $g\left( x \right)$. From the equation of $g\left( x \right)$, we can calculate the value of g\left( 0 \right).

Formula Used:
$\dfrac{d}{{dx}}{e^{mx}} = m{e^{mx}}$
$\dfrac{d}{{dx}}{e^x} = m{e^x}$
$\int {{e^x}dx} = {e^x} + c$

Complete step by step solution:
The given integration is
$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx = g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$
Now we will solve the left side of the equation.
$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx$
$ = \int {\left[ {\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + \left( {{e^x} - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}} \right]} dx$
$ = \int {\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} + \int {\left( {{e^x} - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $
$ = {I_1} + {I_2}$
Now solving ${I_1} = \int {\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $
${I_1} = \int {\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $
Taking common ${e^x}$ from $\left( {{e^{2x}} + {e^x} - 1} \right)$
$ \Rightarrow {I_1} = \int {{e^x}\left( {{e^x} + 1 - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $
Apply the formula ${a^m} \cdot {a^n} = {a^{m + n}}$
$ \Rightarrow {I_1} = \int {\left( {{e^x} + 1 - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}dx} $
Let ${e^x} + {e^{ - x}} + x = z$
Differentiate both sides
$\left( {{e^x} - {e^{ - x}} + 1} \right)dx = dz$
Substitute $\left( {{e^x} - {e^{ - x}} + 1} \right)dx = dz$ and ${e^x} + {e^{ - x}} + x = z$ in ${I_1} = \int {\left( {{e^x} + 1 - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}dx} $
$ \Rightarrow {I_1} = \int {{e^z}dz} $
Applying the formula $\int {{e^x}dx} = {e^x} + c$
$ \Rightarrow {I_1} = {e^z} + {c_1}$
Substitute the value of $z$
$ \Rightarrow {I_1} = {e^{{e^x} + {e^{ - x}} + x}} + {c_1}$
Now solving the integration ${I_2} = \int {\left( {{e^x} - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $
Let ${e^x} + {e^{ - x}} = u$
Differentiate both sides
$\left( {{e^x} - {e^{ - x}}} \right)dx = du$
Substitute ${e^x} + {e^{ - x}} = u$ and $\left( {{e^x} - {e^{ - x}}} \right)dx = du$ in ${I_2} = \int {\left( {{e^x} - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $.
${I_2} = \int {{e^u}du} $
Applying the formula $\int {{e^x}dx} = {e^x} + c$
$ \Rightarrow {I_2} = {e^u} + {c_2}$
Substitute the value of $u$
$ \Rightarrow {I_2} = {e^{\left( {{e^x} + {e^{ - x}}} \right)}} + {c_2}$
Substitute the value of ${I_1}$ and ${I_2}$ in $\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx = {I_1} + {I_2}$
$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx = {e^{\left( {{e^x} + {e^{ - x}} + x} \right)}} + {e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$ Where ${c_1} + {c_2} = c$
Now putting $\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx = {e^{\left( {{e^x} + {e^{ - x}} + x} \right)}} + {e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$ in the given equation $\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx = g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$ and calculating the value of $g\left( x \right)$.
Therefore
${e^{\left( {{e^x} + {e^{ - x}} + x} \right)}} + {e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c = g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$
$ \Rightarrow {e^{\left( {{e^x} + {e^{ - x}}} \right)}} \cdot {e^x} + {e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c = g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$
Now taking common ${e^{\left( {{e^x} + {e^{ - x}}} \right)}}$ from the left side expression
$ \Rightarrow {e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} + 1} \right) + c = g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$
$ \Rightarrow \left( {{e^x} + 1} \right) = g\left( x \right)$
Now putting $x = 0$ in the above equation
$ \Rightarrow \left( {{e^0} + 1} \right) = g\left( 0 \right)$
Since ${a^0} = 1$, so ${e^0} = 1$
$ \Rightarrow \left( {1 + 1} \right) = g\left( 0 \right)$
$ \Rightarrow g\left( 0 \right) = 2$

Option ‘B’ is correct

Note: If an integration is not solved directly, then we will apply the substitution method. In the given question, you need to rewrite the expression under integration so that we can apply the substitution method. By using the substitution method, solve the integration and compare both sides to get the expression of g(x).