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# Evaluate$\int {\dfrac{1}{{({a^2} - {x^2})}}} dx$, where $a > x$

Last updated date: 19th Jul 2024
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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.