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**Hint:**We are given here to integrate the function \[\dfrac{1}{{({a^2} - {x^2})}}\], where \[a > x\]. To do this, first of all we will expand the denominator using the formula of \[({a^2} - {b^2})\]. Then we will multiply and divide the numerator and denominator by \[2a\]. Now we will write the numerator as \[a + x + a - x\]. Now we will solve further and get the value of this integration.

**Formula used:**We use the following formulas here,

\[\ln a - \ln b = \ln \dfrac{a}{b}\]

\[\int {\dfrac{1}{x}dx = \ln x + c} \]

\[({a^2} - {b^2}) = (a + b)(a - b)\]

**Complete step-by-step solution:**

We are given to integrate\[\int {\dfrac{1}{{({a^2} - {x^2})}}} dx\].

We know that \[({a^2} - {b^2}) = (a + b)(a - b)\]. Using this formula, we write the function \[\dfrac{1}{{({a^2} - {x^2})}}\] as,

\[\dfrac{1}{{({a^2} - {x^2})}} = \dfrac{1}{{(a + x)(a - x)}}\]

We multiply and divide the numerator and denominator by \[2a\].

\[ \Rightarrow \dfrac{1}{{({a^2} - {x^2})}} = \dfrac{{2a}}{{2a}} \cdot \dfrac{1}{{(a + x)(a - x)}}\]

We now add and subtract \[x\] on the numerator as,

\[ \Rightarrow \dfrac{1}{{({a^2} - {x^2})}} = \dfrac{{2a}}{{2a}} \cdot \dfrac{{x - x}}{{(a + x)(a - x)}}\]

We will now rearrange this function,

\[ \Rightarrow \dfrac{1}{{({a^2} - {x^2})}} = \dfrac{{(a + x) + (a - x)}}{{2a(a + x)(a - x)}}\]

We will use this step in the integration. We replace \[\dfrac{1}{{({a^2} - {x^2})}}\] with RHS of above step as,

\[\int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\int {\dfrac{{(a + x) + (a - x)}}{{(a + x)(a - x)}}dx} \]

On simplifying further,

\[

\Rightarrow \int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\int {\left( {\dfrac{{a + x}}{{(a + x)(a - x)}}} \right)dx + \dfrac{1}{{2a}}\int {\left( {\dfrac{{a - x}}{{(a + x)(a - x)}}} \right)dx} } \\

\Rightarrow \int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\int {\left( {\dfrac{1}{{(a - x)}}} \right)dx + \dfrac{1}{{2a}}\int {\left( {\dfrac{1}{{(a + x)}}} \right)dx} } \\

\]

We know that \[\int {\dfrac{1}{x}dx = \ln x + c} \], using this above we get,

\[ \Rightarrow \int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\left( { - \ln (a - x)} \right) + \dfrac{1}{{2a}}\left( {\ln \left( {a + x} \right)} \right)\]

Since, \[a > x\] we can safely put the value \[a - x\] in log function as \[a - x > 0\].

\[ \Rightarrow \int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\left[ {\ln \left( {a + x} \right) - \ln (a - x)} \right]\]

As we know that \[\ln a - \ln b = \ln \dfrac{a}{b}\], we move ahead as,

\[ \Rightarrow \int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\left[ {\ln \left( {\dfrac{{a + x}}{{a - x}}} \right)} \right]\]

Hence we have got the value of the given integration as \[\dfrac{1}{{2a}}\left[ {\ln \left( {\dfrac{{a + x}}{{a - x}}} \right)} \right]\].

**Note:**This is to note that we could have also done this integration by the method of partial fraction. But that method is a bit longer. Moreover, if you are not able to find the suitable modifications on the function you can always use a partial fraction method in such types of questions. In partial fraction, we will write the integration as,

\[\int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \int {\left[ {\dfrac{A}{{a + x}} + \dfrac{B}{{a - x}}} \right]dx} \]

We then find the value of \[A\] and \[B\] and then solve ahead.

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