$\displaystyle \int{\dfrac{1}{\sqrt{x^2+a^2}}}\,dx$ Rule

Formula

$\displaystyle \int{\dfrac{1}{\sqrt{x^2+a^2}}}\,dx$ $\,=\,$ $\log_{e}{\big|x+\sqrt{x^2+a^2}\big|}+c$

Introduction

Let, $x$ represents a variable and $a$ denotes a constant. The reciprocal of the square root of sum of squares of them forms a special irrational function in mathematics and its integral with respect to the variable $x$ is written mathematically as follows in calculus.

$\displaystyle \int{\dfrac{1}{\sqrt{x^2+a^2}}}\,dx$

The indefinite integral of one divided by the square root of $x$ square plus $a$ square with respect to $x$ is equal to the natural logarithm of $x$ plus square root of sum of squares of $x$ and $a$, plus the integral constant $c$.

$\therefore\,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{x^2+a^2}}}\,dx$ $\,=\,$ $\ln{\big|x+\sqrt{x^2+a^2}\big|}+c$

Uses

It is used as a formula in finding the integration of a rational function in which the square root of sum of the squares of a variable and a constant appears in the denominator.

Example

Evaluate $\displaystyle \int{\dfrac{1}{\sqrt{x^2+9}}}\,dx$

The variable $x$ is in square form under the square root in the denominator and the number $9$ can also be written in the same form.

$=\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{x^2+3^2}}}\,dx$

The integral of the irrational function is exactly same as the integral rule of multiplicative inverse of the square root of sum of two squares formula. It clears that $a$ is equal to $3$ in this example.

$=\,\,$ $\log_{e}{\big|x+\sqrt{x^2+3^2}\big|}+c$

$=\,\,$ $\log_{e}{\big|x+\sqrt{x^2+9}\big|}+c$

Proof

Learn how to derive an integration rule to find integral of the reciprocal of the square root of sum of the squares of a variable and constant with respect to a variable in mathematics.

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