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# How do you evaluate the integral $\int {\dfrac{1}{{{x^2} - 4}}dx}$?

Last updated date: 13th Jun 2024
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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.