Question

How do you integrate $\dfrac{2x}{{{x}^{2}}-1}$ ?

Answer 1

Hint: Here we have been given a value in fraction and we have to integrate it. We will use the substitution method in this question. Firstly we will let the denominator term equal to some variable and find its differentiation. Then we will replace the value obtained in the original value. Finally we will find the integration of the obtained value and put back the value we left in the final answer and get our desired answer.

Complete step by step solution:
We have to integrate the below value:
$\dfrac{2x}{{{x}^{2}}-1}$
So we will write it as:
$\int{\dfrac{2x}{{{x}^{2}}-1}dx}$….$\left( 1 \right)$
Now we will use substitution method
Let,
${{x}^{2}}-1=u$…$\left( 2 \right)$
Differentiate both sides with respect to $x$
$\Rightarrow \dfrac{d}{dx}{{x}^{2}}-\dfrac{d}{dx}1=\dfrac{d}{dx}u$
Use the formula $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$on left side also differentiation of constant is zero.
$\Rightarrow 2x-0=\dfrac{du}{dx}$
$\Rightarrow 2x=\dfrac{du}{dx}$
Take the denominator value from right side to left side as follows:
$\Rightarrow 2x\,dx=du$…$\left( 3 \right)$
Replace the value from equation (2) and (3) in equation (1) we get,
$\Rightarrow \int{\dfrac{du}{u}}$
Now using the formula $\int{\dfrac{1}{x}dx}=\log x+c$ where $c$is any constant above we get,
$\Rightarrow \log u+c$
Replace the value of $u$ from equation (2) above we get,
$\Rightarrow \log \left( {{x}^{2}}-1 \right)+c$
So we get the value as $\log \left( {{x}^{2}}-1 \right)+c$ .
Hence on integrating the value $\dfrac{2x}{{{x}^{2}}-1}$ we get the answer as $\log \left( {{x}^{2}}-1 \right)+c$ where $c$ is any constant.

Note:
When we are doing integration without the limit we have to always add a constant to our answer. It is done because there are families of functions which differ by only a constant whose anti-derivative is the same function. We can use the partial fraction method also but it becomes quite complicated so if we see that on differentiating the denominator we are getting the numerator we should always use the substitution method.