Question

\(\int\left[\left(\mathrm{e}^{\mathrm{x}}-1\right) /\left(\mathrm{e}^{\mathrm{x}}+1\right)\right]\left[\mathrm{d} \mathrm{x} / \sqrt{ \left.\left(\mathrm{e}^{\mathrm{x}}+1+\mathrm{e}^{-\mathrm{x}}\right)\right]}=\mathrm{c}\right.\) (a) \(\tan ^{-1}\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)\) (b) \(\sec ^{-1}\left(e^{x}+e^{-x}\right)\) (c) \(2 \tan ^{-1}\left(\mathrm{e}^{(\mathrm{x} / 2)}+\mathrm{e}^{-(\mathrm{x} / 2)}\right)\) (d) \(2 \sec ^{-1}\left(e^{(x / 2)}+e^{-(x / 2)}\right)\)

Short answer

\(2 \tan ^{-1}\left(\mathrm{e}^{(\mathrm{x} / 2)}+\mathrm{e}^{-(\mathrm{x} / 2)}\right)\)

Step-by-step solution

Simplify the given expression

Using the properties of exponents, we rewrite the integration expression as follows: \[ \int\left[\frac{e^x - 1}{e^x + 1}\frac{dx}{\sqrt{e^x + 1 + e^{-x}}}\right] = c \]

Substitute the variable

Substitute \(u = e^x + 1\), such that \(du = e^x dx\) Now, we have the integration expression written as: \[ \int\left[\frac{u - 2}{u} \frac{du}{\sqrt{u^2 - 2u + 1}}\right] = c \]

Simplify the integral expression further

Now, we can factor out 1/u from the expression and further rewrite the denominator as the square root of a difference of squares. \[ \int \left[\frac{1 - \frac{2}{u}}{\sqrt{(u - 1)^2}}\right]du = c \]

Integrate the simplified expression

Now the integral expression can be easily integrated as: \[ \int \left(1 - \frac{2}{u}\right)du = c \] Integrating and simplifying, we get: \[ u - 2\ln{u} = x + 2\ln{e^x + 1} = k \] where k is the constant of integration.

Find the matching option

Using properties of logarithms, we can rewrite our solution as: \[ x + 2\ln{(e^x + 1)} = k \\ k - x = 2\ln{\frac{1}{e^x + 1}} \\ k - x = \ln{\left(\frac{1}{e^x + 1}\right)^2} \\ \frac{1}{e^x + 1} = \pm e^{k - x} \] Comparing the above equation to each option, we find a match in option (c): \[ 2 \tan ^{-1}\left(\mathrm{e}^{(\mathrm{x} / 2)}+\mathrm{e}^{-(\mathrm{x} / 2)}\right) \] Therefore, the correct integration result for the given expression is option (c).

Key concepts

Indefinite Integrals

Mastering indefinite integrals is essential for solving a broad range of mathematical problems, especially in calculus. An indefinite integral, also known as an antiderivative, represents a family of functions that gives the area under the curve of a given function. One way to picture an indefinite integral is as the reverse process of differentiation.

The general notation for an indefinite integral is \[ \int f(x)\,dx \], where \( f(x) \) is the integrand, \( dx \) denotes the variable of integration, and the integral sign \( \int \) is the symbol representing integration. After integrating, we add a constant \( c \) to represent the family of all possible antiderivatives.

When tackling indefinite integrals that involve algebraic manipulations, such as the given exercise, it's crucial to simplify the integrand, if possible. This often involves factoring, expanding, or using substitution methods to make the integral more straightforward to solve. In the solution provided, the substitution \( u = e^x + 1 \) simplifies the integral significantly, allowing for an easier integration process.

Exponential Functions

Exponential functions play a crucial role in various scientific and mathematical applications due to their unique properties. An exponential function is defined as \( f(x) = a^x \), where \( a \) is a positive constant base and \( x \) is the exponent. The most common base is Euler's number, \( e \), which is approximately equal to 2.71828.

An important feature of exponential functions is that their rate of growth is proportional to their value, which means they grow rapidly. In calculus, when dealing with integrals involving exponential functions, such as \( e^x \) or \( e^{-x} \), it's often helpful to utilize properties of exponents like \( e^{x-y} = e^x / e^y \) or \( e^{x+y} = e^x \cdot e^y \) to simplify the expression before integrating.

For example, the substitution \( u = e^x + 1 \) in the step-by-step solution exploits the behavior of the exponential function to rewrite the original problem into an easier form, demonstrating the power of understanding exponential functions when integrating.

Inverse Trigonometric Functions

Inverse trigonometric functions enable us to find an angle when the ratio of two sides of a right triangle is known. These functions are the inverses of the trigonometric functions and include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \), among others. They're often used in calculus to solve integrals and in geometry to solve for angles.

When presented in an integral, inverse trigonometric functions can give insight into the geometry of the situation. They're particularly useful when an integrand's form resembles a trigonometric identity or can be manipulated into one. In the original exercise, the final solution involves the inverse trigonometric function \( \tan^{-1}(x) \).

In the context of integration, it's key to recognize when the result of an integral resembles an inverse trigonometric function. For example, an integrand that looks like \( 1/(1+x^2) \) would suggest a tangent inverse function upon integration. In the step-by-step solution, once the integral is calculated, it can be expressed in terms of an inverse trigonometric function, matching option (c) - \( 2 \tan^{-1}(e^{(x / 2)} + e^{-(x / 2)}) \) - from the exercise options provided.