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

The integral of reciprocal of the square root of difference of squares is equal to the sum of natural logarithm of sum of variable and square root of difference of squares, and the constant of integration.

Let $x$ be a variable and $a$ be a constant. The square root of the difference of their squares forms an irrational function in the following form.

$\sqrt{x^2-a^2}$

In some cases, this irrational function also appears in reciprocal form as follows.

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

The indefinite integral of this type of irrational function in reciprocal form with respect to $x$ is written as follows.

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

The integral of one by square root of $x$ squared minus $a$ squared with respect to $x$ is called the integral rule for the square root of difference of squares in reciprocal form.

The integral of this irrational function is equal to the natural logarithm of $x$ plus square root of $x$ squared minus $a$ squared, and plus integral constant $c$.

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

According to the representation of natural logarithms, the natural logarithm can also be written in following form in mathematics.

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

Evaluate $\displaystyle \int{\dfrac{1}{\sqrt{x^2-16}}}\,dx$

In this example problem, the number $16$ can be expressed in square form.

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{x^2-4^2}}}\,dx$

Take $a \,=\, 4$ and substitute it in the integral rule for the square root of difference of squares in reciprocal form.

$=\,\,\,$ $\log_{e}{\Big|x+\sqrt{x^2-4^2}\Big|}$ $+$ $c$

$=\,\,\,$ $\log_{e}{\Big|x+\sqrt{x^2-16}\Big|}$ $+$ $c$

Learn how to derive the integral rule for the square root of difference of squares in reciprocal form.

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