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**Hint:**We solve this using partial fraction. We can split \[{x^2} - 4\] using the algebraic identity \[{a^2} - {b^2} = (a + b)(a - b)\]. Since 4 can be expressed as 2, raise to the power 2. That is \[4 = {2^2}\]. After that we simplify by partial fraction then we apply the sum integral rule to solve the given problem. The term inside the integral sign is called the integrand.

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

Given \[\int {\dfrac{1}{{{x^2} - 4}}dx} \]

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

Then \[{x^2} - 4\] can be written as \[{x^2} - {2^2}\]

Applying the identity we have,

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

\[\Rightarrow \int {\dfrac{1}{{{x^2} - 4}}dx} = \int {\dfrac{1}{{(x + 2)(x - 2)}}dx} \]

Now take the integrand and applying the partial fraction,

\[\Rightarrow \dfrac{1}{{(x + 2)(x - 2)}} = \dfrac{A}{{(x + 2)}} + \dfrac{B}{{(x - 2)}}\]

Taking LCM and simplifying we have,

\[\Rightarrow 1 = A(x - 2) + B(x + 2)\]

Now we need to find ‘A’ and ‘B’.

Put \[x = 2\]in above equation,

\[\Rightarrow 1 = A(2 - 2) + B(2 + 2)\]

\[\Rightarrow 1 = 4B\]

Rearranging we have

\[\Rightarrow 4B = 1\]

Divide the whole equation by 4,

\[\Rightarrow B = \dfrac{1}{4}\]

Now put \[x = - 2\], we have

\[\Rightarrow 1 = A( - 2 - 2) + B( - 2 + 2)\]

\[\Rightarrow 1 = - 4A\]

Rearranging we have

\[\Rightarrow - 4A = 1\]

Divide the whole equation by -4,

\[\Rightarrow A = - \dfrac{1}{4}\]

Then \[\dfrac{1}{{(x + 2)(x - 2)}} = \dfrac{A}{{(x + 2)}} + \dfrac{B}{{(x - 2)}}\]becomes

\[\dfrac{1}{{(x + 2)(x - 2)}} = - \dfrac{1}{4}\dfrac{1}{{(x + 2)}} + \dfrac{1}{4}\dfrac{1}{{(x - 2)}}\]

Now applying the integration we have

\[\Rightarrow \int {\dfrac{1}{{(x + 2)(x - 2)}}dx} = \int { - \dfrac{1}{4}\dfrac{1}{{(x + 2)}}} dx + \int {\dfrac{1}{4}\dfrac{1}{{(x - 2)}}dx} \]

\[\Rightarrow \int { - \dfrac{1}{4}\dfrac{1}{{(x + 2)}}} dx + \int {\dfrac{1}{4}\dfrac{1}{{(x - 2)}}dx} \]

Taking the constant term outside the integral sign

\[\Rightarrow - \dfrac{1}{4}\int {\dfrac{1}{{(x + 2)}}} dx + \dfrac{1}{4}\int {\dfrac{1}{{(x - 2)}}dx} \]

We know that \[\int {\dfrac{1}{x}dx = \log x + c} \]. In each term in above equation that is \[(x + 2)\] and \[\Rightarrow (x - 2)\] as ‘x’ term and we integrate it,

\[\Rightarrow - \dfrac{1}{4}\log |x + 2| + \dfrac{1}{4}\log |x - 2| + c\]

Where ‘c’ is the integration constant.

Rearranging we have,

\[\Rightarrow \dfrac{1}{4}\left( {\log |x - 2| - \log |x + 2|} \right) + c\]

We know the logarithm law that is \[\log \left( {\dfrac{m}{n}} \right) = \log (m) - \log (n)\], applying this we have:

\[\Rightarrow \dfrac{1}{4}\log \left( {\dfrac{{x - 2}}{{x + 2}}} \right) + c\].

Thus we have

\[\int {\dfrac{1}{{{x^2} - 4}}dx} = \dfrac{1}{4}\log \left( {\dfrac{{x - 2}}{{x + 2}}} \right) + c\], where ‘c’ is the integration constant.

**Note:**In the given above problem we have an indefinite integral, that is no upper and lower limit. Hence we add the integration constant ‘c’ after integrating. In a definite integral we will have an upper and lower limit, we don’t need to add integration constant in the case of definite integral. We have different integration rule:

The power rule: If we have a variable ‘x’ raised to a power ‘n’ then the integration is given by \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} \].

The constant coefficient rule: if we have an indefinite integral of \[K.f(x)\], where f(x) is some function and ‘K’ represent a constant then the integration is equal to the indefinite integral of f(x) multiplied by ‘K’. That is \[\int {K.f(x)dx = c\int {f(x)dx} } \].

The sum rule: if we have to integrate functions that are the sum of several terms, then we need to integrate each term in the sum separately. That is

\[\int {\left( {f(x) + g(x)} \right)dx = \int {f(x)dx} } + \int {g(x)dx} \]

For the difference rule we have to integrate each term in the integrand separately.

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