Question

Solve $\int {{e^{{e^{{e^x}}}}}} \cdot {e^{{e^x}}} \cdot {e^x}dx = ... + C$
A.${e^{{e^x}}}$
B.$\left( {\dfrac{1}{2}} \right){e^2}{e^{{e^x}}}$
C.${e^{{e^{{e^x}}}}}$
D.$\left( {\dfrac{1}{2}} \right){e^{{e^x}}}$

Answer 1

Hint : Here in this question given an Indefinite integral, we have to find the integrated value of a given exponential function. This can be solved by the substitution method and later integrated by using the standard exponential formula of integration. And by further simplification we get the required solution.

Complete step-by-step answer :
Integration is the inverse process of differentiation. An integral which does not have any upper and lower limit is known as an indefinite integral.
Consider the given function.
$\Rightarrow \,\int {{e^{{e^{{e^x}}}}}} \cdot {e^{{e^x}}} \cdot {e^x}dx$
By the property of exponent ${a^m} \cdot {a^n} = {a^{m + n}}$
It can be rewritten as:
$\Rightarrow \,\int {{e^{{e^{{e^x}}} + {e^x} + x}}} dx$-------(1)
Let us take substitution
$u = {e^x}$
Differentiate with respect to x
$\dfrac{{du}}{{dx}} = {e^x}$
$dx = {e^{ - x}}du$
Then equation (1) becomes
$\Rightarrow \,\int {{e^{{e^u} + u + x}} \cdot {e^{ - x}}du}$
$\Rightarrow \,\int {{e^{{e^u} + u + x - x}}du}$
$\Rightarrow \,\int {{e^{{e^u} + u}}du}$ -----(2)
Again, take substitution
$v = {e^u}$
Differentiate with respect to u
$\dfrac{{dv}}{{du}} = {e^u}$
$du = {e^{ - u}}dv$
Then equation (2) becomes
$\Rightarrow \,\int {{e^{v + u}} \cdot {e^{ - u}}dv}$
$\Rightarrow \,\int {{e^{v + u - u}}dv}$
$\Rightarrow \,\int {{e^v}\,\,dv}$
On integrating, we get
$\Rightarrow \,\,{e^v} + C$
put $v = {e^u}$, then we have
$\Rightarrow \,\,{e^{{e^u}}} + C$
put $u = {e^x}$, then we have
$\Rightarrow \,\,{e^{{e^{{e^x}}}}} + C$
Where, $C$ is an integrating constant.
$\therefore \,\,\int {{e^{{e^{{e^x}}}}}} \cdot {e^{{e^x}}} \cdot {e^x}dx = \,\,{e^{{e^{{e^x}}}}} + C$
Hence, it’s a required solution.
Therefore, Option (C) is the correct answer.
So, the correct answer is “Option C”.

Note : By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. The standard integration formulas for the exponential function must be known.