Question

Find $\int{\dfrac{{{x}^{e-1}}+{{e}^{x-1}}}{{{x}^{e}}+{{e}^{x}}}dx}$ is equal to
(a) $\log \left( {{x}^{e}}+{{e}^{x}} \right)+c$
(b) $e\log \left( {{x}^{e}}+{{e}^{x}} \right)+c$
(c) $\dfrac{1}{e}\log \left( {{x}^{e}}+{{e}^{x}} \right)+c$
(d) None of the above

Answer 1

In this type of question we have to use the concept of integration. We know that $\int{\dfrac{f'\left( x \right)}{f\left( x \right)}dx=\log \left( f\left( x \right) \right)}$ so here we try to adjust the derivative of the denominator in the numerator and then by applying this formula we obtain the required result.

Complete step-by-step solution:

Now, we have to find the value of $\int{\dfrac{{{x}^{e-1}}+{{e}^{x-1}}}{{{x}^{e}}+{{e}^{x}}}dx}$
Let us consider,
$\Rightarrow \int{\dfrac{{{x}^{e-1}}+{{e}^{x-1}}}{{{x}^{e}}+{{e}^{x}}}dx}$
We know that, $\int{\dfrac{f'\left( x \right)}{f\left( x \right)}dx=\log \left( f\left( x \right) \right)}$ so we are going to apply this rule of integration here.
Let us adjust derivative of denominator in the numerator
$\Rightarrow \int{\dfrac{{{x}^{e-1}}+{{e}^{x-1}}}{{{x}^{e}}+{{e}^{x}}}dx}=\int{\dfrac{{{x}^{e-1}}+\dfrac{{{e}^{x}}}{e}}{{{x}^{e}}+{{e}^{x}}}dx}$
Now we simplify the numerator and hence we get,

$\Rightarrow \int{\dfrac{{{x}^{e-1}}+{{e}^{x-1}}}{{{x}^{e}}+{{e}^{x}}}dx}=\int{\dfrac{e{{x}^{e-1}}+{{e}^{x}}}{e\left( {{x}^{e}}+{{e}^{x}} \right)}}dx$
As we know that $e$ is the constant so by taking $\dfrac{1}{e}$ outside the integration sign we get,
$\Rightarrow \int{\dfrac{{{x}^{e-1}}+{{e}^{x-1}}}{{{x}^{e}}+{{e}^{x}}}dx}=\dfrac{1}{e}\int{\dfrac{e{{x}^{e-1}}+{{e}^{x}}}{\left( {{x}^{e}}+{{e}^{x}} \right)}}dx$
Also we know that, $\dfrac{d}{dx}\left( {{x}^{e}}+{{e}^{x}} \right)=e{{x}^{e-1}}+{{e}^{x}}$. Hence, we can write
$\Rightarrow \int{\dfrac{{{x}^{e-1}}+{{e}^{x-1}}}{{{x}^{e}}+{{e}^{x}}}dx}=\dfrac{1}{e}\int{\dfrac{\dfrac{d}{dx}\left( {{x}^{e}}+{{e}^{x}} \right)}{\left( {{x}^{e}}+{{e}^{x}} \right)}}dx$
Now, by applying $\int{\dfrac{f'\left( x \right)}{f\left( x \right)}dx=\log \left( f\left( x \right) \right)}$ we can write,
$\Rightarrow \int{\dfrac{{{x}^{e-1}}+{{e}^{x-1}}}{{{x}^{e}}+{{e}^{x}}}dx}=\dfrac{1}{e}\log \left( {{x}^{e}}+{{e}^{x}} \right)+c$
Hence, option (c) is the correct option.

Note: In this type of question students have to apply rules of integration. Students have to remember that integration is the inverse process of differentiation. Also students have to note that there are two types of definite integral and indefinite integral. Definite integral has both upper and lower limits so that after evaluating definite integral students will get a finite value. Indefinite integral does not have defined upper and lower limits but after evaluating indefinite integral students have to add $c$ to the final answer which is called a constant of integration.