Problem 45 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{d x}{x\left(x^{2}-1\right)^{3 / 2}}, x>1$$

Step-by-Step Solution

Answer

In summary, the indefinite integral of the given expression is: $$\int \frac{d x}{x\left(x^{2}-1\right)^{3 / 2}} = 2\left(\ln{\left|\sqrt{x^2 - 1}\right|} - \frac{\sqrt{x^2 - 1}}{2x} + \frac{1}{4}\ln{\left|\frac{x-1}{x+1}\right|}\right) + C$$

Step 1: Select an appropriate substitution

To simplify the integral, we can use the following substitution based on the factor \(x^2 - 1\): $$x = \cosh{t}$$ With this substitution, we will be able to rewrite the integral in terms of the hyperbolic cosine function.

Step 2: Find the derivative of the substitution

Now, we need to find the derivative of the substitution, \(dx\) in terms of \(dt\): $$\frac{dx}{dt} = \sinh{t}$$ $$dx = \sinh{t}\,dt$$

Step 3: Substitute and rewrite the integral

We substitute the expressions for \(x\) and \(dx\) into the integral: $$\int \frac{d x}{x\left(x^{2}-1\right)^{3 / 2}} = \int \frac{\sinh{t}\,dt}{\cosh{t}(\cosh^2{t}-1)^{3 / 2}}$$ Now, we simplify the integral using the identity \(\cosh^2{t} - \sinh^2{t} = 1\): $$\int \frac{\sinh{t}\,dt}{\cosh{t}(\cosh^2{t}-1)^{3 / 2}} = \int \frac{\sinh{t}\,dt}{\cosh{t}(\sinh^2{t})^{3 / 2}}$$ Simplify further: $$\int \frac{\sinh{t}\,dt}{\cosh{t}(\sinh^2{t})^{3 / 2}} = \int \frac{\sinh{t}\,dt}{\cosh{t}\sinh^{3}{t}} = \int \frac{dt}{\cosh{t}\sinh^{2}{t}}$$

Step 4: Evaluate the simplified integral

The integral is now in a simpler form: $$\int \frac{dt}{\cosh{t}\sinh^{2}{t}}$$ Notice that \(\cosh{t}\sinh^{2}{t} = \frac{1}{2}\left(\cosh^2{t} - 1\right)\sinh{t} = \frac{1}{2}\left(\sinh^2{t} - \sinh^4{t}\right)\) The integral becomes: $$\int \frac{dt}{\frac{1}{2}\left(\sinh^2{t} - \sinh^4{t}\right)} = 2\int \frac{dt}{\sinh^2{t} - \sinh^4{t}}$$ Now, let's set \(u = \sinh{t}\) $$\int \frac{dt}{\sinh^2{t} - \sinh^4{t}} = \int \frac{du}{u^2 - u^4} = \int \frac{du}{u^2(1-u^2)}$$

Step 5: Perform partial fraction decomposition

To compute the integral, we perform a partial fraction decomposition: $$\frac{1}{u^2(1-u^2)} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{1-u} + \frac{D}{1+u} $$ Solving for A, B, C, and D, we find: $$A = 1, B = -\frac{1}{2}, C = \frac{1}{4}, D = \frac{1}{4}$$ Now, we can rewrite the integral as: $$2\int \frac{du}{u^2 - u^4} = 2\int \left(\frac{1}{u} - \frac{1}{2u^2} + \frac{1}{4(1-u)} + \frac{1}{4(1+u)}\right)du$$

Step 6: Evaluate the integral

Now, we can evaluate the integral: $$2\int \frac{du}{u^2 - u^4} = 2\left(\ln{|u|} - \frac{1}{2}u^{-1} + \frac{1}{4}\ln{|1-u|} - \frac{1}{4}\ln{|1+u|}\right) + C$$

Step 7: Reverse the substitution

Now we reverse the substitution \(u = \sinh{t}\) and \(x = \cosh{t}\): $$2\left(\ln{|\sinh{t}|} - \frac{1}{2}\sinh^{-1}{t} + \frac{1}{4}\ln{|1-\sinh{t}|} - \frac{1}{4}\ln{|1+\sinh{t}|}\right) + C$$ Finally, substitute \(t\) with \(\operatorname{arccosh}(x)\): $$2\left(\ln{|\sinh(\operatorname{arccosh}(x))|} - \frac{1}{2}\sinh^{-1}(\sinh(\operatorname{arccosh}(x))) + \frac{1}{4}\ln{|1-\sinh(\operatorname{arccosh}(x))|} - \frac{1}{4}\ln{|1+\sinh(\operatorname{arccosh}(x))|}\right) + C$$ So the answer is: $$2\left(\ln{\left|\sqrt{x^2 - 1}\right|} - \frac{\sqrt{x^2 - 1}}{2x} + \frac{1}{4}\ln{\left|\frac{x-1}{x+1}\right|}\right) + C$$

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving integrals that are otherwise difficult to evaluate directly. This method involves changing the variable of integration to transform a complex integral into a simpler form.
This can be particularly helpful when an integral contains expressions that resemble derivatives of known functions or function identities, such as trigonometric or hyperbolic identities.

In practice, when using the substitution method, you:

• Select an appropriate substitution, choosing a new variable that simplifies the integral.
• Rewrite the integral in terms of this new variable.
• Integrate with respect to the new variable.
• Finally, reverse the substitution back to the original variable.

In the given exercise, the substitution \(x = \cosh{t}\) enabled the transformation of the original integral into one involving hyperbolic functions, thereby simplifying the evaluation process.

Hyperbolic Functions

Hyperbolic functions, similar to their trigonometric counterparts, are functions of an angle that appear frequently in calculus and complex analysis. They include:

• Sinh, or hyperbolic sine, \(\sinh{x}\), defined as \(\sinh{x} = \frac{e^x - e^{-x}}{2}\).
• Cosh, or hyperbolic cosine, \(\cosh{x}\), defined as \(\cosh{x} = \frac{e^x + e^{-x}}{2}\).

These functions exhibit properties similar to trigonometric functions such as identities and derivatives, which make them particularly useful for calculus.
In particular, they satisfy the identity \(\cosh^2{x} - \sinh^2{x} = 1\).

In the problem exercise, hyperbolic functions were used to format the integral in a way that allowed direct application of trigonometric integration techniques. Through \(\cosh{t}\) and \(\sinh{t}\), the substitution \(x = \cosh{t}\) enabled a smooth transformation into a form suitable for further simplification.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to express a rational function as a sum of simpler fractions, which can then be integrated individually. This method is extremely useful when dealing with complex rational functions that are not straightforward to integrate.
When using partial fraction decomposition, ensure that:

• The degree of the numerator is less than the degree of the denominator.
• If necessary, perform long division to satisfy the condition above.

Then, decompose the function into fractions with linear or quadratic terms in the denominators.

In the step-by-step solution, partial fraction decomposition was applied to the expression \(\frac{1}{u^2(1-u^2)}\) to break it down into simpler fractions. The task involved finding constants \(A\), \(B\), \(C\), and \(D\) such that the overall expression could be split into terms that are easy to integrate, resulting in a smooth solution for the integral.