Question

Evaluate: \[\int {\dfrac{{{e^{x - 1}} + {x^{e - 1}}}}{{{e^x} + {x^e}}}} dx\]

Answer 1

Hint:
Here we are using the concept of Integration. Integration is the way of adding slices to find the whole. Integration is a method of summation along the data. We will use the integration by substitution to simplify the integral and simplifying it further we will get the required answer.

Formula Used:
We will use the following formulas:
1) Product Rule: \[{a^x} \cdot {a^y} = {a^{x + y}}\]
2) Differentiation Formula: \[\dfrac{{d({e^x})}}{{dx}} = {e^x}\] and \[\dfrac{{d({x^{e - 1}})}}{{dx}} = e{x^{e - 1}}\]
3) Integration Formula: \[\begin{array}{l}\int {\dfrac{1}{x}} dx = \log x + C\\\end{array}\] where \[\dfrac{1}{\begin{array}{l}x\\\end{array}}\] is called the Integrand and C is an arbitrary constant introduced at the end of an Integration.

Complete step by step solution:
In Integration by Substitution, usually numerator will be the derivative of the denominator. So we assume the denominator to be a variable.
Let us take \[{e^x} + {x^e} = t\] .
\[ \Rightarrow t = {e^x} + {x^e}\]
Differentiating with respect to\[x\] on both the sides using the a differentiation Formula\[\dfrac{{d({e^x})}}{{dx}} = {e^x}\] and \[\dfrac{{d({x^{e - 1}})}}{{dx}} = e{x^{e - 1}}\] , we get
\[ \Rightarrow \dfrac{{dt}}{{dx}} = {e^x} + e{x^{e - 1}}\]
\[ \Rightarrow dt = {e^x} + e{x^{e - 1}}dx\]
Dividing by \[e\] on both the sides, we get
\[ \Rightarrow \dfrac{{dt}}{e} = \dfrac{1}{e}\left( {{e^x} + e{x^{e - 1}}} \right)dx\]
\[ \Rightarrow \dfrac{{dt}}{e} = \left( {\dfrac{{{e^x}}}{e} + \dfrac{{e{x^{e - 1}}}}{e}} \right)dx\]
Now using the Product Rule \[{a^x} \cdot {a^y} = {a^{x + y}}\] to simplify the above equation, we get
\[ \Rightarrow \dfrac{1}{e}dt = \left( {{e^x}.{e^{ - 1}} + {x^{e - 1}}} \right)dx\]
\[ \Rightarrow \dfrac{1}{e}dt = \left( {{e^{x - 1}} + {x^{e - 1}}} \right)dx\]
Substituting \[{e^{x - 1}} + {x^{e - 1}}dx = \dfrac{{dt}}{e}\] and \[{e^x} + {x^e} = t\] , we get
\[ \Rightarrow \int {\dfrac{{{e^{x - 1}} + {x^{e - 1}}}}{{{e^x} + {x^e}}}} dx = \int {\dfrac{1}{t}} \dfrac{{dt}}{e}\]
Using the Integration Formula, \[\begin{array}{l}\int {\dfrac{1}{x}} dx = \log x + C\\\end{array}\] and taking \[\dfrac{1}{e}\] as common , we get
\[ \Rightarrow \dfrac{1}{e}\int {\dfrac{1}{t}} dt = \dfrac{1}{e}\log t + C\]
Now, again by Substituting \[t = {e^x} + {x^e}\] , we get
\[ \Rightarrow \dfrac{1}{e}\int {\dfrac{1}{t}} dt = \dfrac{1}{e}.\log \left( {{e^x} + {x^e}} \right) + C\]

Therefore, \[\int {\dfrac{{{e^{x - 1}} + {x^{e - 1}}}}{{{e^x} + {x^e}}}} dx = \dfrac{{\log ({e^x} + {x^e})}}{e} + C\].

Note:
Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. By doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function. We can use this method to find an integral value when it is set up in the special form. It means that the given integral is of the form: \[\int {f\left( {g(x)} \right)} .g'(x).dx = f(u).du\] . Here, first we will integrate the function with respect to the substituted value (\[f\left( u \right)\]), and finish the process by substituting the original function, \[g\left( x \right)\]. It is the counterpart to the chain rule for differentiation, in fact, it can loosely be thought of as using the chain rule "backwards".