Question

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

Answer 1

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.