Question

\(I=\int[1 /\\{(x-4) \sqrt{(x-2)\\}}] d x=f(x)+c\) then \(f(x)\) is (a) \((1 / \sqrt{2}) \log |[\\{\sqrt{(x-2)}-\sqrt{2}\\} /\\{\sqrt{(x-2)}+\sqrt{2}\\}]|+c\) (b) \((1 / 2) \log \mid[\\{\sqrt{x}-2)-\sqrt{2}\\} /\\{\sqrt{(x-2)}+\sqrt{2}\\}] \mid+c\) (c) \((1 / 2 \sqrt{2}) \log \mid[\\{\sqrt{(x-2)}+\sqrt{2}\\} /\\{\sqrt{(x-2)}-\sqrt{2}\\}]+c\) (d) none of these

Short answer

The short answer is: \(f(x) = \frac{1}{\sqrt{2}} \log \left|\frac{\sqrt{x-2}-\sqrt{2}}{\sqrt{x-2}+\sqrt{2}}\right|+c\). The correct option is (a).

Step-by-step solution

Identify the integral to compute

Firstly, identify the given integral which is \(I=\int[1 /\\{(x-4) \sqrt{(x-2)\\}}] d x\). We need to find the antiderivative function \(f(x)\) of this integral.

Apply substitution

As it could be quite difficult to tackle the original integral directly. This will simplify the given integral for easier computation. Here we could use substitution method. Let \(u=\sqrt{x-2}\), then \(du=1/(2\sqrt{x-2}) dx\) and on rearranging, get \(dx = 2\sqrt{x-2}du\). Notice that \(x = u^2 + 2\) and \(x-4 = u^2 - 2\) then change all the values into \(u\) on the main equation and simplify.

Compute the integral using standard integral formulas

Now the integral in terms of \(u\) is \(\int[1 /\\{(u^2 -2) u\\}] 2u du = 2 \int[1/(u^2 -2)du]\). This integral can be computed using the standard integral formula that states that \(\int[1/x] dx = \log |x|+c\). After integrating, we get \(2\log |u -\sqrt{2}| - 2\log |u + \sqrt{2}|+c'\) where \(c'\) is a constant.

Apply reverse substitution

Now replace \(u\) by \(\sqrt{x-2}\) and divide by \(\sqrt{2}\) to simplify into the form as options: \(1/\sqrt{2} [\log |\sqrt{x-2} -\sqrt{2}| - \log |\sqrt{x-2} + \sqrt{2}|]+c\).

Match the simplified integral with the provided options

It clearly matches with option (a). Hence, \(f(x) = 1/\sqrt{2} [\log |(\sqrt{x-2}-\sqrt{2}) / (\sqrt{x-2}+\sqrt{2})|]+c \). So, the correct answer is option (a).

Key concepts

Indefinite Integrals

When faced with an indefinite integral, you're basically being asked to find the antiderivative of a function. These types of integrals are not tied to definite limits of integration and include a constant of integration, generally denoted as 'c'. An indefinite integral is usually signified as \( \int f(x) dx = F(x) + c \) where \( F(x) \) represents the antiderivative of \( f(x) \) and \( c \) is the constant. The process of finding this \( F(x) \) is known as anti-differentiation or integration.

It's crucial to understand that the result you get after integrating is a family of functions that all differentiate to the original function \( f(x) \) due to the addition of the constant 'c'. Integrals without clear bounds are useful for representing general solutions to differential equations or for setting up the integration for further manipulation, which might involve finding a definite integral or applying additional techniques.

Substitution Method

The substitution method is a powerful tool for solving more complex indefinite integrals. It works by transforming the integral into a simpler form, which is easier to evaluate. Think of it as a make-over for mathematical expressions! The crux is to identify a part of the integral as a new variable \( u \), such that \( du \) (the derivative of \( u \) with respect to \( x \) times \( dx \)) simplifies the integral.

Once a suitable substitution is made, the integral is rewritten in terms of \( u \), which then ideally resembles a function with a known integral. After integrating with respect to \( u \), you then 'back-substitute' the original variable \( x \) in place of \( u \) to obtain the solution expressed in terms of the original variable.

Antiderivatives

An antiderivative of a function \( f(x) \) is a function whose derivative yields \( f(x) \). In simpler words, if you're given \( F(x) \) as the antiderivative of \( f(x) \) then, \( F'(x) = f(x) \). The process of finding antiderivatives is also known as integration, which is the inverse operation of differentiation.

The solution to an indefinite integral consists of all antiderivatives, which is why we include the constant 'c' representing an infinite number of functions that differ only by a constant value. Antiderivatives play a critical role in solving differential equations and are used to predict the behavior of systems described by those equations.

Integral Formulas

Integral formulas are nothing short of a gold mine for anyone venturing into the realm of integration! They are predefined antiderivatives of common functions that are used to simplify the integration process. For instance, knowing that the integral of \( 1/x \) is \( \log |x| + c \) allows you to integrate functions that, upon simplification, resemble the form \( 1/x \) with ease.

Memorizing the basic integral formulas is often recommended as it makes the process of integrating much faster and serves as a foundation for understanding more advanced integration techniques. These formulas are the bread and butter of integrals and can be stitched together creatively to solve more intricate problems.