Question

Evaluate the following integrals: $$ \int \frac{d x}{x \sqrt{\left(x^{4}-1\right)}} $$

Short answer

Answer: The result of the integral is $-\frac{\sqrt{x^4 - 1}}{x^2} + C$.

Step-by-step solution

Substitute with hyperbolic trigonometric functions

The most suitable substitution for this exercise is replacing \(x\) with the inverse hyperbolic cosine function, denoted by \(\cosh{u}\). This substitution is useful because of how it simplifies the square root of the quartic polynomial. $$ x = \cosh{u} $$

Calculate dx and simplify the integral

We need to find the derivative of \(x\) with respect to \(u\) and replace \(dx\) in the integral. We have: $$ \frac{dx}{du} = \sinh{u} $$ Now, we can replace \(dx\) with \(du\): $$ dx = \sinh{u}\, du $$ Substitute \(x\) and \(dx\) into the integral: $$ \int \frac{d x}{x \sqrt{\left(x^{4}-1\right)}} = \int \frac{\sinh{u}\, du}{\cosh{u} \sqrt{\left(\cosh^4{u}-1\right)}} $$

Simplify the square root

One advantage of using the hyperbolic cosine function is that it has a simple identity regarding the difference of squares: $$ \cosh^2{u} - \sinh^2{u} = 1 $$ With this identity, the quartic polynomial in the square root can now be simplified: $$ \cosh^4{u} - 1 = (\cosh^2{u} + 1)(\cosh^2{u} - 1) = \sinh^2{u}\cosh^2{u} $$ Now, substitute this simplification of the quartic polynomial back into the integral: $$ \int \frac{\sinh{u}\, du}{\cosh{u} \sqrt{\sinh^2{u}\cosh^2{u}}} $$

Continue simplifying the integral

Now, we can simplify the integral by canceling some terms and taking square roots. We obtain: $$ \int \frac{\sinh{u}\, du}{\cosh{u} \sqrt{\sinh^2{u}\cosh^2{u}}} = \int \frac{\sinh{u}\, du}{\cosh{u} \cdot \sinh{u}\cosh{u}} $$ Cancel out \(\sinh{u}\) and \(\cosh{u}\) in the numerator and denominator: $$ \int \frac{du}{\cosh^{2}{u}} $$

Evaluate the integral

Now, we can evaluate the integral by finding the antiderivative of the simplified function: $$ \int \frac{du}{\cosh^{2}{u}} = \int \operatorname{sech}^{2}{u}\, du = -\tanh{u} + C $$ Where \(C\) is the constant of integration.

Substitute back the original variable

The final step is to substitute back the original variable \(x\) using the initial substitution \(x = \cosh{u}\). Since \(\tanh{u} = \frac{\sinh{u}}{\cosh{u}}\), we can write the following relationship: $$ \tanh{u} = \frac{\sqrt{x^4 - 1}}{x^2} $$ Then we have: $$ -\tanh{u} + C = -\frac{\sqrt{x^4 - 1}}{x^2} + C $$

Final answer

Thus, the final answer is: $$ \int \frac{d x}{x \sqrt{\left(x^{4}-1\right)}} = -\frac{\sqrt{x^4 - 1}}{x^2} + C $$