How to Evaluate the Integral ∫ 2/ (4 - x^4) dx with Hyperbolic Functions and Special Techniques

In this article, we will explore the process of evaluating the integral ∫ 2/ (4 - x^4) dx using advanced mathematical techniques involving hyperbolic functions, specifically the inverse hyperbolic cotangent function (coth^{-1} x). This detailed guide will not only provide a clear solution but also explain the necessary steps and transformations.

Understanding the Hyperbolic Inverse Cotangent Function

The hyperbolic inverse cotangent function, denoted as coth^{-1} x, has a special role in evaluating certain integrals. This function is derived from the hyperbolic cotangent function, which is closely related to the inverse functions of certain exponential expressions. For a given x, the hyperbolic cotangent function is defined as:

cot theta x

Using a series of mathematical transformations, we can express the inverse function in terms of x.

Deriving the Inverse Hyperbolic Cotangent Function

Starting from the definition, we have:

cot theta x
left(frac{e^{theta} e^{-theta}}{e^{theta} - e^{-theta}}right) x

Applying the componendo-dividendo rule, we simplify the expression:

left(frac{e^{theta} e^{-theta}}{e^{theta} - e^{-theta}} cdot frac{e^{theta} - e^{-theta}}{e^{theta} e^{-theta}}right) left(frac{x 1}{x - 1}right)

Further manipulation yields:

left(frac{e^{2theta}}{2}right) left(frac{x 1}{x - 1}right)

Solving for θ, we get:

2theta lnleft(frac{x 1}{x - 1}right)
theta frac{1}{2}lnleft(frac{x 1}{x - 1}right)

Thus, the inverse hyperbolic cotangent function can be expressed as:

arccot x frac{1}{2}lnleft(frac{x 1}{x - 1}right)
arccotleft(frac{x^2}{2}right) frac{1}{2}lnleft(frac{left(frac{x^2}{2}right) 1}{left(frac{x^2}{2}right) - 1}right)
frac{1}{2}lnleft(frac{x^2 2}{x^2 - 2}right)

Evaluating the Integral

To evaluate the integral int frac{2x}{4 - x^4} dx, we first rewrite it in a more convenient form:

int frac{2x}{4 - x^4} dx int frac{2x}{2^2 - x^4} dx
frac{1}{2}int frac{du}{2 - u^2} dx
where u x^2
frac{1}{4}int frac{du}{2 - u^2}

Using the integral formula for the inverse hyperbolic tangent function, we get:

int frac{du}{a^2 - u^2} frac{1}{2a}lnleft(frac{a u}{a - u}right) C

Substituting the values, we have:

frac{1}{4}int frac{du}{2 - u^2} frac{1}{4} cdot frac{1}{2}lnleft(frac{sqrt{2} u}{sqrt{2} - u}right) C

Finally, substituting back u x^2, we obtain:

frac{1}{4} cdot frac{1}{2}lnleft(frac{sqrt{2} x^2}{sqrt{2} - x^2}right) C frac{1}{4}lnleft(frac{x^2 2}{x^2 - 2}right) C

Therefore, the integral is:

int frac{2x}{4 - x^4} dx frac{1}{2}arccotleft(frac{x^2}{2}right) C

Conclusion

This step-by-step guide demonstrates how to evaluate the integral ∫ 2/ (4 - x^4) dx using the inverse hyperbolic cotangent function. The key steps involve transforming the integral into a more manageable form and applying the appropriate integral formula. This approach not only solves the problem but also provides insight into the power of hyperbolic functions in integration.