Question

Find the value of integral $\int\limits_{0}^{1}{\dfrac{dx}{{{e}^{x}}+e}}$ .

Answer 1

Hint: Start by simplification of the definite integral using the property $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( 1-x \right)dx}$ . After simplification, let the denominator of the simplified function be t. Solve the integral, don’t forget to change the limits while putting the t.

Complete step-by-step solution -
Before starting the solution, let us discuss the important properties of definite integration.
Some important properties are:
$\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$
$\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{c}{f\left( x \right)dx+}\int\limits_{c}^{b}{f\left( x \right)dx}$
Now let us start with the integral given in the above question.
$\int\limits_{0}^{1}{\dfrac{dx}{{{e}^{x}}+e}}$
Using the property $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$ , we get
$\int\limits_{0}^{1}{\dfrac{dx}{{{e}^{0+1-x}}+e}}$
$=\int\limits_{0}^{1}{\dfrac{dx}{\dfrac{e}{{{e}^{x}}}+e}}$
$=\int\limits_{0}^{1}{\dfrac{{{e}^{x}}dx}{e\left( 1+{{e}^{x}} \right)}}$
Now to convert the integral to a form from where we can directly integrate it,
let ( $1+{{e}^{x}}$ ) = t.
Now differentiate both side w.r.t x
[As we know the derivative of $e^x$ =$e^x$ ]
$\therefore {{e}^{x}}dx=dt$
So, if we substitute the terms in the integral in terms of t and also limits in terms of t. So, lower limit at x=0 is ( $1+{e}^0$ ) = 2 and upper limit at x=1 is ( $1+{e}^1$ ) =( $1+e$ ), Hence our integral becomes
$=\int\limits_{2}^{1+e}{\dfrac{dt}{et}}$
Now we know that $\int{\dfrac{dx}{x}}$ is equal to lnx +c. So, our integral comes out to be:
\[=\left[ \dfrac{\ln t}{e} \right]_{2}^{1+e}\]
\[=\dfrac{\ln \left( 1+e \right)-\ln 2}{e}\]
Now we know that $\ln a-\ln b=\ln \left( \dfrac{a}{b} \right)$ . using this in our expression, we get
\[=\dfrac{1}{e}\ln \left( \dfrac{1+e}{2} \right)\]
So, we can say that the value of $\int\limits_{0}^{1}{\dfrac{dx}{{{e}^{x}}+e}}$ is equal to \[\dfrac{1}{e}\ln \left( \dfrac{1+e}{2} \right)\] .

Note: Remember that in case of definite integral whenever you try to express a function in terms of some other variable, the limits also change. The general mistake a student makes is they substitute x but forget to change the limits leading to errors. Also, you need to remember all the basic formulas that we use for indefinite integrals as they are used in definite integrations as well.