If x > a, then \[\int{\dfrac{dx}{{{x}^{2}}-{{a}^{2}}}}=\]?

Hint: This type of question is based on the concept of integration. First we have to simplify the given function using the formula \[{{x}^{2}}-{{a}^{2}}=\left( x+a \right)\left( x-a \right)\] in the denominator. Then, we need to multiply the numerator and denominator by 2a. Express the numerator as 2a=(x+a)-(x-a). Use the rule \[\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}\] to simplify the function further. Then substitute u=x+a and v=x-a. Integrate the functions separately with respect to u and v. substitute back the functions in terms of x to the final answer after integration.

Complete step by step answer:
According to the question, we are asked to find \[\int{\dfrac{dx}{{{x}^{2}}-{{a}^{2}}}}\].
We have been given the function is \[\dfrac{1}{{{x}^{2}}-{{a}^{2}}}\]. --------(1)
We know that \[{{x}^{2}}-{{a}^{2}}=\left( x+a \right)\left( x-a \right)\].
Using this formula in the function (1), we get
\[\dfrac{1}{{{x}^{2}}-{{a}^{2}}}=\dfrac{1}{\left( x+a \right)\left( x-a \right)}\]
Now, let us divide the numerator and denominator by 2a.
\[\Rightarrow \dfrac{1}{{{x}^{2}}-{{a}^{2}}}=\dfrac{2a}{\left( x+a \right)\left( x-a \right)\times 2a}\]
We can write the function as \[\dfrac{1}{{{x}^{2}}-{{a}^{2}}}=\dfrac{1}{2a}\times \dfrac{2a}{\left( x+a \right)\left( x-a \right)}\].
Now, let us add and subtract x in the numerator. We get
\[\dfrac{1}{{{x}^{2}}-{{a}^{2}}}=\dfrac{1}{2a}\times \dfrac{2a+x-x}{\left( x+a \right)\left( x-a \right)}\]
\[\Rightarrow \dfrac{1}{{{x}^{2}}-{{a}^{2}}}=\dfrac{1}{2a}\times \dfrac{a+a+x-x}{\left( x+a \right)\left( x-a \right)}\]
On arranging the terms in the numerator, we get
\[\dfrac{1}{{{x}^{2}}-{{a}^{2}}}=\dfrac{1}{2a}\times \dfrac{x+a-x+a}{\left( x+a \right)\left( x-a \right)}\]
Let us take -1 common from the last two terms of the numerator.
\[\Rightarrow \dfrac{1}{{{x}^{2}}-{{a}^{2}}}=\dfrac{1}{2a}\times \dfrac{x+a-\left( x-a \right)}{\left( x+a \right)\left( x-a \right)}\]
We know that \[\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}\]. Using this rule of division in the above function, we get
\[\Rightarrow \dfrac{1}{{{x}^{2}}-{{a}^{2}}}=\dfrac{1}{2a}\left[ \dfrac{x+a}{\left( x+a \right)\left( x-a \right)}-\dfrac{x-a}{\left( x+a \right)\left( x-a \right)} \right]\]
We find that x+a is common in the first fraction. Cancelling out x+a. We get
\[\dfrac{1}{{{x}^{2}}-{{a}^{2}}}=\dfrac{1}{2a}\left[ \dfrac{1}{\left( x-a \right)}-\dfrac{x-a}{\left( x+a \right)\left( x-a \right)} \right]\]
Now, x-a is common in the second fraction of the function. We get
\[\dfrac{1}{{{x}^{2}}-{{a}^{2}}}=\dfrac{1}{2a}\left[ \dfrac{1}{\left( x-a \right)}-\dfrac{1}{\left( x+a \right)} \right]\] --------------(2)
We need to integrate \[\dfrac{1}{{{x}^{2}}-{{a}^{2}}}\] with respect to x.
\[\Rightarrow \int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx}=\int{\dfrac{1}{2a}\left[ \dfrac{1}{\left( x-a \right)}-\dfrac{1}{\left( x+a \right)} \right]}dx\]
We find that \[\dfrac{1}{2a}\] is a constant.
Therefore, we get
\[\int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{1}{2a}\int{\left[ \dfrac{1}{\left( x-a \right)}-\dfrac{1}{\left( x+a \right)} \right]}dx\]
Using the subtraction rule of integration, we get
\[\int{\left( u-v \right)dx=\int{udx-\int{vdx}}}\]
We get
\[\int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{1}{2a}\left[ \int{\dfrac{1}{\left( x-a \right)}}dx-\int{\dfrac{1}{\left( x+a \right)}dx} \right]\] ---------(3)
Let us integrate \[\int{\dfrac{1}{\left( x-a \right)}}dx\] first.
Let u=x-a
Differentiating u with respect to x, we get
\[\dfrac{du}{dx}=\dfrac{d}{dx}\left( x-a \right)\]
\[\Rightarrow \dfrac{du}{dx}=\dfrac{dx}{dx}-\dfrac{d}{dx}\left( a \right)\]
We know that \[\dfrac{dx}{dx}=1\] and differentiation of a constant is 0.
Therefore, we get
\[\dfrac{du}{dx}=1\]
\[\therefore du=dx\]
\[\Rightarrow \int{\dfrac{1}{\left( x-a \right)}}dx=\int{\dfrac{1}{u}}du\]
We know that \[\int{\dfrac{1}{x}}dx=\log x+c\]. Using this rule of integration, we get
\[\int{\dfrac{1}{u}}du=\log u+{{c}_{1}}\]
But we know that u=x-a.
Therefore, \[\int{\dfrac{1}{\left( x-a \right)}}dx=\log \left( x-a \right)+{{c}_{1}}\].
Now, consider \[\int{\dfrac{1}{\left( x+a \right)}}dx\].
Let v=x+a
Differentiating v with respect to x, we get
\[\dfrac{dv}{dx}=\dfrac{d}{dx}\left( x+a \right)\]
\[\Rightarrow \dfrac{dv}{dx}=\dfrac{dx}{dx}+\dfrac{d}{dx}\left( a \right)\]
We know that \[\dfrac{dx}{dx}=1\] and differentiation of a constant is 0.
Therefore, we get
\[\dfrac{dv}{dx}=1\]
\[\therefore dv=dx\]
\[\Rightarrow \int{\dfrac{1}{\left( x+a \right)}}dx=\int{\dfrac{1}{v}}dv\]
We know that \[\int{\dfrac{1}{x}}dx=\log x+c\]. Using this rule of integration, we get
\[\int{\dfrac{1}{v}}dv=\log v+{{c}_{2}}\]
But we know that v=x+a.
Therefore, \[\int{\dfrac{1}{\left( x+a \right)}}dx=\log \left( x+a \right)+{{c}_{2}}\].
Substitute this integration in equation (3).
\[\Rightarrow \int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{1}{2a}\left[ \log \left( x-a \right)+{{c}_{1}}-\log \left( x+a \right)-{{c}_{2}} \right]\]
Let us take the constants outside the bracket.
\[\Rightarrow \int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{1}{2a}\left[ \log \left( x-a \right)-\log \left( x+a \right) \right]+\dfrac{{{c}_{1}}-{{c}_{2}}}{2a}\]
Let us consider \[k=\dfrac{{{c}_{1}}-{{c}_{2}}}{2a}\], where k is a constant.
Therefore, we get
\[\int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{1}{2a}\left[ \log \left( x-a \right)-\log \left( x+a \right) \right]+k\]
We know that \[\log a-\log b=\log \dfrac{a}{b}\].
Using this rule of logarithm in the integration, we get
\[\int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{1}{2a}\left[ \log \dfrac{\left( x-a \right)}{\left( x+a \right)} \right]+k\]
\[\therefore \int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{1}{2a}\log \dfrac{\left( x-a \right)}{\left( x+a \right)}+k\]
Therefore, the integration of \[\dfrac{1}{{{x}^{2}}-{{a}^{2}}}\] with respect to x is \[\dfrac{1}{2a}\log \dfrac{\left( x-a \right)}{\left( x+a \right)}+k\].

Note:
Whenever we get this type of question, we have to simplify the given function to a simplified form for easy integration. We have to know that integration of \[\dfrac{1}{x}\] is log x. Also be thorough with the rules and properties of logarithm. Avoid calculation mistakes based on sign convention.