Math Doubts

Integral rule for Reciprocal of difference of squares

Formula

$\displaystyle \int{\dfrac{1}{x^2-a^2}\,}dx \,=\, \dfrac{1}{2a}\log_e{\Bigg|\dfrac{x-a}{x+a}\Bigg|}+c$

Introduction

Let $x$ represents a variable and $a$ represents a constant. The subtraction of square of $a$ from $x$ squared is written as $x^2-a^2$ in mathematics. The multiplicative inverse of the difference of them is written in the following mathematical form.

$\dfrac{1}{x^2-a^2}$

The indefinite integral of this rational expression with respect to $x$ is written in calculus as follows.

$\implies$ $\displaystyle \int{\dfrac{1}{x^2-a^2}\,}dx$

The indefinite integration of this rational function is equal to product of the reciprocal of the two times the constant and the natural logarithm of the quotient of the difference by sum of them.

$\implies$ $\displaystyle \int{\dfrac{1}{x^2-a^2}\,}dx \,=\, \dfrac{1}{2a}\ln{\Bigg|\dfrac{x-a}{x+a}\Bigg|}+c$

Alternative forms

The indefinite integral formula for multiplicative inverse of the difference of squares can be written in terms of any two literals.

$(1) \,\,\,$ $\displaystyle \int{\dfrac{1}{y^2-m^2}\,}dy \,=\, \dfrac{1}{2m}\log_e{\Bigg|\dfrac{y-m}{y+m}\Bigg|}+c$

$(2) \,\,\,$ $\displaystyle \int{\dfrac{1}{l^2-d^2}\,}dl \,=\, \dfrac{1}{2d}\log_e{\Bigg|\dfrac{l-d}{l+d}\Bigg|}+c$

$(3) \,\,\,$ $\displaystyle \int{\dfrac{1}{z^2-f^2}\,}dz \,=\, \dfrac{1}{2f}\log_e{\Bigg|\dfrac{z-f}{z+f}\Bigg|}+c$

Example

Evaluate $\displaystyle \int{\dfrac{1}{x^2-3^2}\,}dx$

Take $a = 3$ and substitute it in the integral property.

$=\,\,\,$ $\dfrac{1}{2 \times 3}\log_e{\Bigg|\dfrac{x-3}{x+3}\Bigg|}+c$

$=\,\,\,$ $\dfrac{1}{6}\log_e{\Bigg|\dfrac{x-3}{x+3}\Bigg|}+c$

Proof

Learn how to derive the indefinite integration rule for multiplicative inverse of difference of squares.

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