Solving the Integral ( int frac{sqrt{1 x^2}}{x} dx ) Using Various Substitution Methods

Introduction

This article explores the integral of the function ( int frac{sqrt{1 x^2}}{x} dx ) and provides several methods for solving it. We will use algebraic substitution, trigonometric substitution, and algebraic manipulation techniques to derive the solution step-by-step.

Method 1: Algebraic Substitution

Let's start by breaking down the integral into more manageable parts through algebraic manipulation.

( int frac{sqrt{1 x^2}}{x} dx int frac{1 - x^2 1}{x sqrt{1 x^2}} dx int frac{1}{x sqrt{1 x^2}} dx - int frac{x^2}{x sqrt{1 x^2}} dx int frac{1 - x^2^{-1/2} 2x}{x^2} dx - int frac{1}{2} frac{1}{sqrt{1 x^2}^2}dx ) - int frac{1 x^2^{-1/2} 2x}{1 - sqrt{1 x^2}^2} d(sqrt{1 x^2}) - tanh^{-1} sqrt{1 x^2} sqrt{1 x^2} - frac{1}{2} ln left| frac{sqrt{1 x^2} - 1}{sqrt{1 x^2} 1} right| C )

Method 2: Trigonometric Substitution

Next, we employ a trigonometric substitution. Let ( 1 x^2 z^2 ), which implies ( 2x dx 2z dz ) and thus ( dx frac{z dz}{x} ).

So, the integral becomes:

( I int frac{sqrt{1 x^2}}{x} dx int frac{z}{x^2} frac{z dz}{x} int frac{z^2}{x^2} dz int frac{z^2}{z^2 - 1} dz z - frac{1}{2} ln left| frac{z - 1}{z 1} right| C sqrt{1 x^2} - frac{1}{2} ln left| frac{sqrt{1 x^2} - 1}{sqrt{1 x^2} 1} right| C )

Method 3: Algebraic Manipulation

Let us now consider another approach by letting ( 1 x^2 y^2 ), which implies ( x^2 1 - y^2 ) and ( x dx -y dy ). The integral now becomes:

( I int frac{sqrt{1 x^2}}{x^2} dx -int frac{sqrt{y^2}}{y^2 - 1} frac{-y dy}{y} int frac{y^2 - 1 1}{y^2 - 1} dy int 1 dy - int frac{1}{y^2 - 1} dy y - frac{1}{2} ln left| frac{y - 1}{y 1} right| C sqrt{1 x^2} - frac{1}{2} ln left| frac{sqrt{1 x^2} - 1}{sqrt{1 x^2} 1} right| C )

Conclusion

These methods illustrate different ways to solve the integral ( int frac{sqrt{1 x^2}}{x} dx ). Each approach showcases the power of algebraic manipulation, trigonometric substitution, and substitution methods, which are commonly used in solving complex integrals.