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Question: How do you integrate \(\dfrac{2x}{{{x}^{2}}-1}\) ?...

How do you integrate 2xx2−1\dfrac{2x}{{{x}^{2}}-1} ?

Explanation

Solution

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:
2xx2−1\dfrac{2x}{{{x}^{2}}-1}
So we will write it as:
∫2xx2−1dx\int{\dfrac{2x}{{{x}^{2}}-1}dx}….(1)\left( 1 \right)
Now we will use substitution method
Let,
x2−1=u{{x}^{2}}-1=u…(2)\left( 2 \right)
Differentiate both sides with respect to xx
⇒ddxx2−ddx1=ddxu\Rightarrow \dfrac{d}{dx}{{x}^{2}}-\dfrac{d}{dx}1=\dfrac{d}{dx}u
Use the formula ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}on left side also differentiation of constant is zero.
⇒2x−0=dudx\Rightarrow 2x-0=\dfrac{du}{dx}
⇒2x=dudx\Rightarrow 2x=\dfrac{du}{dx}
Take the denominator value from right side to left side as follows:
⇒2x dx=du\Rightarrow 2x\,dx=du…(3)\left( 3 \right)
Replace the value from equation (2) and (3) in equation (1) we get,
⇒∫duu\Rightarrow \int{\dfrac{du}{u}}
Now using the formula ∫1xdx=log⁡x+c\int{\dfrac{1}{x}dx}=\log x+c where ccis any constant above we get,
⇒log⁡u+c\Rightarrow \log u+c
Replace the value of uu from equation (2) above we get,
⇒log⁡(x2−1)+c\Rightarrow \log \left( {{x}^{2}}-1 \right)+c
So we get the value as log⁡(x2−1)+c\log \left( {{x}^{2}}-1 \right)+c .
Hence on integrating the value 2xx2−1\dfrac{2x}{{{x}^{2}}-1} we get the answer as log⁡(x2−1)+c\log \left( {{x}^{2}}-1 \right)+c where cc 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.