How to Integrate 1/(4x^2 - 225) Using Different Methods

Integration is a fundamental concept in calculus, where derivatives and integrals are inverse operations. This article will explore the process of integrating the expression 1/(4x^2 - 225) using multiple techniques. The provided solutions cover trigonometric substitution, partial function decomposition, and a table of integrals.

Method 1: Trigonometric Substitution

One straightforward approach to integrating 1/(4x^2 - 225) is to use trigonometric substitution. Let's break down this process:

Substitute the variable: x frac{5}{2}tantheta Then, dx frac{5}{2}sec^2theta, dtheta Using the identity, tan^2theta 1 sec^2theta, we can simplify the integral as follows:

begin{align*} int frac{1}{4x^2 - 225} , dx int frac{1}{4left(frac{5}{2}tanthetaright)^2 - 225} cdot frac{5}{2}sec^2theta, dtheta int frac{frac{5}{2}sec^2theta}{25tan^2theta - 225} , dtheta int frac{frac{5}{2}sec^2theta}{25(tan^2theta - 9)} , dtheta int frac{frac{5}{2}sec^2theta}{25(sec^2theta - 10)} , dtheta int frac{frac{5}{2}}{25(sec^2theta - 10)} , dtheta int frac{1}{10(sec^2theta - 10)} , dtheta frac{1}{10} int frac{1}{sec^2theta - 10} , dtheta frac{1}{10} int frac{1}{1 tan^2theta - 10} , dtheta frac{1}{10} int frac{1}{tan^2theta - 9} , dtheta frac{1}{10} int frac{1}{9(frac{1}{9}tan^2theta - 1)} , dtheta frac{1}{90} int frac{1}{(frac{1}{3}tantheta)^2 - 1} , d(3theta) frac{1}{90} cdot frac{3}{2} arctanleft(frac{1}{3}tanthetaright) C frac{1}{60} arctanleft(frac{1}{3}tanthetaright) C frac{1}{60} arctanleft(frac{2x}{5}right) C end{align*}

Method 2: Using a Table of Integrals

A more straightforward method involves referencing a table of integrals:

int frac{1}{4x^2 - 225} , dx int frac{1}{4(x^2 - frac{225}{4})} , dx frac{1}{4} int frac{1}{x^2 - (frac{15}{2})^2} , dx

Using the standard integral form, int frac{1}{x^2 - a^2} , dx frac{1}{a} arctanleft(frac{x}{a}right) C, we get:

int frac{1}{4x^2 - 225} , dx frac{1}{4} cdot frac{1}{frac{15}{2}} arctanleft(frac{x}{frac{15}{2}}right) C frac{2x}{30} arctanleft(frac{2x}{15}right) C frac{1}{15} arctanleft(frac{2x}{15}right) C

Method 3: Decomposition Using Partial Fractions

Another way to solve this integral is by decomposing it into partial fractions:

int frac{1}{4x^2 - 225} , dx int frac{1}{4(x^2 - frac{225}{4})} , dx frac{1}{4} int frac{1}{(2x - 15)(2x 15)} , dx

Using partial fraction decomposition, we write:

frac{1}{(2x - 15)(2x 15)} frac{A}{2x - 15} frac{B}{2x 15}

Solving for A and B, we find A frac{1}{30} and B -frac{1}{30}. Hence:

begin{align*} int frac{1}{4x^2 - 225} , dx frac{1}{4} int left(frac{-1}{30(2x 15)} frac{1}{30(2x - 15)}right) , dx frac{1}{4} left(-frac{1}{30} int frac{1}{2x 15} , dx frac{1}{30} int frac{1}{2x - 15} , dxright) frac{1}{4} left(-frac{1}{60} ln|2x 15| frac{1}{60} ln|2x - 15|right) C frac{1}{240} lnleft|frac{2x - 15}{2x 15}right| C end{align*}

Conclusion

In conclusion, there are various methods to integrate the expression 1/(4x^2 - 225). Each method offers a unique insight into the problem. Whether using trigonometric substitution, referencing a table of integrals, or partial fraction decomposition, each approach provides a pathway to the solution. Understanding the underlying concepts and techniques can greatly enhance one's problem-solving skills in calculus and beyond.