Question

What is the value of the integral $\int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx$ ?
(A) $- \left( {\dfrac{{{e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \right) + c$
(B) $- \left( {{e^{2x}} + 1} \right) + c$
(C) ${\left( {\dfrac{1}{{{e^x} + 1}}} \right)^2} + c$
(D) ${\left( {\dfrac{1}{{{e^x} + {e^{ - x}}}}} \right)^2} + c$

Answer 1

The given question requires us to integrate a function of x with respect to x. We evaluate the given integral using the substitution method. We substitute the exponential function as t and then follow the substitution method of integration. After getting the answer in the terms of t, we substitute back t in terms of x.

Complete step by step answer:
The given question requires us to find the value of the indefinite integral $\int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx$.
So, we have, $\int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx$
But, it is very difficult to integrate the function directly. So, we can assign a new variable to the exponential function ${e^x}$.
So, let $t = {e^x} - - - - - \left( 1 \right)$.
Then, differentiating both sides of the equation, we get,
$\Rightarrow dt = {e^x}dx - - - - - \left( 2 \right)$
So, the integral $\int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx$ can be simplified by substituting the value of $\left( {dx} \right)$ in terms of $\left( {dt} \right)$ as obtained above. So, we get,
$\int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} \left( {\dfrac{{dt}}{{{e^x}}}} \right)$
Substituting ${e^x}$ as t, we get,
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = \int {\dfrac{2}{{{{\left( {t + \dfrac{1}{t}} \right)}^2}}}} \left( {\dfrac{{dt}}{t}} \right)$
Evaluating the whole square using the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$. So, we have,
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = \int {\dfrac{2}{{\left( {{t^2} + \dfrac{1}{{{t^2}}} + 2} \right)}}} \left( {\dfrac{{dt}}{t}} \right)$
Opening the brackets, we get,
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = \int {\dfrac{{2dt}}{{\left( {{t^3} + \dfrac{1}{t} + 2t} \right)}}}$
Taking LCM in the numerator, we get,
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = \int {\dfrac{{2dt}}{{\left( {\dfrac{{{t^4} + 1 + 2{t^2}}}{t}} \right)}}}$
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = \int {\dfrac{{2tdt}}{{\left( {{t^4} + 1 + 2{t^2}} \right)}}}$
Now, condensing the denominator using algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, we get,
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = \int {\dfrac{{2tdt}}{{{{\left( {{t^2} + 1} \right)}^2}}}}$
Now, we again substitute $u = \left( {{t^2} + 1} \right)$.
Differentiating both sides, we get,
$du = 2tdt$
So, we get,
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = \int {\dfrac{{2tdt}}{{{{\left( {{t^2} + 1} \right)}^2}}}} = \int {\dfrac{{du}}{{{u^2}}}}$
Now, we know the power rule of integration as $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$. So, we get,
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = \int {{u^{ - 2}}du}$
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = \dfrac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}} + c$
Simplifying further, we get,
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = - u + c$
Now, we have the final answer. We just need to substitute u in terms of t and then in terms of x.
So, we get,
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = - \left( {{t^2} + 1} \right) + c$
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = - \left( {{{\left( {{e^x}} \right)}^2} + 1} \right) + c$
Simplifying further, we get,
$\Rightarrow \int {\dfrac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx = - \left( {{e^{2x}} + 1} \right) + c$, where c is any arbitrary constant.

Therefore, option (B) is the correct answer.

Note:
The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the arbitrary constant. It is extremely important to substitute back the value of the assumed variable, once we have reached the final stage.