Question

Completing the square Evaluate the following integrals. $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}}$$

Short answer

$ Answer: $\mathrm{arcoth}(2) - \mathrm{arcoth}(\sqrt{2})$

Step-by-step solution

Complete the square in the denominator expression

First, we rewrite the expression inside the square root as a perfect square. Notice that the product \((x-1)(x-3)\) can be rewritten as \((x^2 - 4x +3)\). Now, let's complete the square: \((x^2 - 4x +3) = (x^2 - 4x + 4) - 1 = (x-2)^2 - 1\) Therefore, the integral becomes: $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-2)^2 - 1}}$$

Perform substitution

Now, we perform the substitution \(u = \frac{x-2}{\sqrt{1}}\) or \(u = x-2\). Then the differential is simply \(du = dx\). The new integration limits for \(u\) are \(u_1 = 2+\sqrt{2} - 2 = \sqrt{2}\) and \(u_2 = 4-2 = 2\). Plugging this into the integral, we get: $$\int_{\sqrt{2}}^{2} \frac{d u}{\sqrt{u^2 - 1}}$$

Integrate using hyperbolic functions

Recall that the antiderivative of \(\frac{1}{\sqrt{u^2 - 1}}\) is \(\mathrm{arcoth}(u) + C\). Therefore, now we can evaluate the integral as follows: $$\int_{\sqrt{2}}^{2} \frac{d u}{\sqrt{u^2 - 1}} = \mathrm{arcoth}(2) - \mathrm{arcoth}(\sqrt{2}) $$

Final result

Thus, the final result of the integral is: $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}} = \mathrm{arcoth}(2) - \mathrm{arcoth}(\sqrt{2})$$

Key concepts

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, are the reverse of differentiation. When we integrate a function, we are essentially finding all the possible functions that, when differentiated, would give the original function.

An indefinite integral is represented by the integral sign followed by the function and the differential, for example, \[ \int f(x) \, dx. \] The 'dx' signifies that we are integrating with respect to the variable 'x'. There is no specific range or limits to the integration, hence it's called indefinite. A constant of integration, usually denoted as 'C', is added because there could be an infinite number of possible antiderivatives each differing by a constant.

Hyperbolic Functions

Hyperbolic functions are analogues of trigonometric functions but for a hyperbola rather than a circle. The primary hyperbolic functions include hyperbolic sine (sinh), hyperbolic cosine (cosh), and hyperbolic tangent (tanh), along with their inverses.

Just like the antiderivative of \(1/x\) is \(\textrm{ln}|x| + C\), one of the antiderivatives for \(1/\sqrt{u^2 - 1}\) is \(\textrm{arcoth}(u) + C\), where 'arcoth' is the inverse hyperbolic cotangent function. These functions can be expressed using exponential functions: for instance, \(\textrm{sinh}(x) = (e^x - e^{-x})/2\). In integration, hyperbolic functions can simplify the process, especially when the integrand involves expressions akin to the forms of these functions.

Integration Techniques

There are various techniques to solve integrals, and choosing the right one depends on the form of the function being integrated. Some common methods include substitution (also known as u-substitution), integration by parts, partial fractions, and trigonometric substitution.

Each technique has its place depending on how the integrand is structured. For instance, u-substitution is often used when an integrand includes a function and its derivative, whereas integration by parts is useful when the integrand is a product of two functions. Completing the square can be particularly helpful for integrating rational functions with quadratic forms in the denominator, as we've seen in the example problem.

U-Substitution

U-substitution is a method used to simplify the integration process. The goal is to transform the integral into a simpler form by changing the variable of integration. It is similar in spirit to applying a change of variables in an algebraic expression to make it easier to solve.

For instance, given an integral involving \(x\), we would choose a substitution \(u = g(x)\) that simplifies the integral, finding \(du/dx\) to replace \(dx\) with \(du\). If the original problem limits are numeric, you'll need to compute the new limits by plugging the original limits into the \(u\)-substitution function.

The integral in the given exercise has been made simpler by utilizing a u-substitution, wherein the square root expression was converted into a perfect square, thereby reducing the integrand into a form that's directly integrable using hyperbolic functions.