Question

Find the value of $\int {\dfrac{{dx}}{{{e^x} + {e^{ - x}}}}}$.

Answer 1

We can rewrite the given expression by substituting ${e^{ - x}} = \dfrac{1}{{{e^x}}}$. Simplifying the expression, we can use the substitution method for integration. We can substitute a new variable for ${e^x}$. Expressing all in terms of $t$, we get a new function. Then we can apply the derivative of ${\tan ^{ - 1}}x$. Substituting back for $t$ we get the answer.

Useful formula:
For any variable $x$ we have, $\dfrac{{d{e^x}}}{{dx}} = {e^x}$
For any variable $t$ we have,
$\dfrac{d}{{dt}}({\tan ^{ - 1}}t) = \dfrac{1}{{1 + {t^2}}}$
$\int {\dfrac{{dt}}{{{t^2} + 1}}} = {\tan ^{ - 1}}t + C$, where $C$ is the constant of integration.

Complete step-by-step answer:
Consider $\dfrac{1}{{{e^x} + {e^{ - x}}}}$
We know, ${e^{ - x}} = \dfrac{1}{{{e^x}}}$.
So we can write the expression as,
$\dfrac{1}{{{e^x} + {e^{ - x}}}} = \dfrac{1}{{{e^x} + \dfrac{1}{{{e^x}}}}}$
Simplifying the expression we get,
$\dfrac{1}{{{e^x} + {e^{ - x}}}} = \dfrac{1}{{\dfrac{{{e^{2x}} + 1}}{{{e^x}}}}}$
$\Rightarrow \dfrac{1}{{{e^x} + {e^{ - x}}}} = \dfrac{{{e^x}}}{{{e^{2x}} + 1}}$
Integrating both sides with respect to $x$ we get,
$\int {\dfrac{{dx}}{{{e^x} + {e^{ - x}}}}} = \int {\dfrac{{{e^x}dx}}{{{e^{2x}} + 1}}}$
We can integrate this using substitution method.
Here we can observe that ${e^{2x}} = {({e^x})^2}$
Now put ${e^x} = t$
Differentiating both sides we get,
${e^x}dx = dt$, since derivative of ${e^x}$ is ${e^x}$ itself.
Also ${e^{2x}} = {({e^x})^2} = {t^2}$
This gives $\dfrac{{{e^x}dx}}{{{e^{2x}} + 1}} = \dfrac{{dt}}{{{t^2} + 1}}$
So we have,
$\int {\dfrac{{dx}}{{{e^x} + {e^{ - x}}}}} = \int {\dfrac{{dt}}{{{t^2} + 1}}}$
We have $\dfrac{d}{{dt}}({\tan ^{ - 1}}t) = \dfrac{1}{{1 + {t^2}}}$
We know that the derivative of a constant is always $0$. So if there is a constant term added to ${\tan ^{ - 1}}t$, the derivative will remain $0$.
This gives $\int {\dfrac{{dt}}{{{t^2} + 1}}} = {\tan ^{ - 1}}t + C$, where $C$ is the constant of integration.
$\Rightarrow \int {\dfrac{{dx}}{{{e^x} + {e^{ - x}}}}} = {\tan ^{ - 1}}t + C$, where $C$ is the constant of integration
Substituting for $t$ we get,
$\Rightarrow \int {\dfrac{{dx}}{{{e^x} + {e^{ - x}}}}} = {\tan ^{ - 1}}({e^x}) + C$, where $C$ is the constant of integration
$\therefore$ The answer is ${\tan ^{ - 1}}({e^x}) + C$, where $C$ is the constant of integration.

Note: We need to rewrite the equation so that we can apply the method of substitution. In the method of substitution, we have to express every term in the function in terms of the new variable used. It is important to add the constant of integration since the integral is indefinite.