Solveeit Logo
Login

Question

Question: How do you integrate \(\left[ {\dfrac{{{x^2} + 1}}{{{x^2} - 1}}} \right] dx\) ?...

How do you integrate [x2+1x2−1]dx\left[ {\dfrac{{{x^2} + 1}}{{{x^2} - 1}}} \right] dx ?

Explanation

Solution

Hint : In the given question, we are required to integrate a rational function involving the second degree of the variable x. There can be various ways to integrate a given function and there may be different answers to an indefinite integral problem. However, all the solutions if done with correct concepts and logic in mind are correct and are inter-convertible into one another.

Complete step by step solution:
In the problem given to us, we have to integrate the function in variable x given to us, [x2+1x2−1]\left[ {\dfrac{{{x^2} + 1}}{{{x^2} - 1}}} \right] with respect to x.
So, ∫[x2+1x2−1] dx\int {\left[ {\dfrac{{{x^2} + 1}}{{{x^2} - 1}}} \right] } \,dx
We cannot directly integrate the given rational function in x because the degree of numerator is the same as the degree of the denominator. So, we first either divide the numerator by denominator or manipulate the given function so as to convert it into a form that is easy to integrate.
Adding and subtracting 11 from numerator and separating the fractions, we get,
⇒∫[(x2−1)+2x2−1] dx\Rightarrow \int {\left[ {\dfrac{{\left( {{x^2} - 1} \right) + 2}}{{{x^2} - 1}}} \right] } \,dx
⇒∫[1+(2x2−1)] dx\Rightarrow \int {\left[ {1 + \left( {\dfrac{2}{{{x^2} - 1}}} \right)} \right] } \,dx
Separating the two integrals, we get,
⇒∫(1) dx+∫(2x2−1)dx\Rightarrow \int {\left( 1 \right)} \,dx + \int {\left( {\dfrac{2}{{{x^2} - 1}}} \right)dx}
We know that integration of 11 with respect to x is x. So, we get,
⇒x+2∫(1x2−12)dx\Rightarrow x + 2\int {\left( {\dfrac{1}{{{x^2} - {1^2}}}} \right)dx}
Now, evaluating ∫(1x2−12)dx\int {\left( {\dfrac{1}{{{x^2} - {1^2}}}} \right)dx} using special integral of the form ∫(1x2−a2)dx=12alog⁡∣x−ax+a∣\int {\left( {\dfrac{1}{{{x^2} - {a^2}}}} \right)dx = \dfrac{1}{{2a}}\log \left| {\dfrac{{x - a}}{{x + a}}} \right|} , we get,
⇒x+2×(12(1)log⁡∣x−1x+1∣)\Rightarrow x + 2 \times \left( {\dfrac{1}{{2\left( 1 \right)}}\log \left| {\dfrac{{x - 1}}{{x + 1}}} \right|} \right)
⇒x+log⁡∣x−1x+1∣+c\Rightarrow x + \log \left| {\dfrac{{x - 1}}{{x + 1}}} \right| + c
So, the integral of the function [x2+1x2−1]\left[ {\dfrac{{{x^2} + 1}}{{{x^2} - 1}}} \right] with respect to x is x+log⁡∣x−1x+1∣+cx + \log \left| {\dfrac{{x - 1}}{{x + 1}}} \right| + c
So, the correct answer is “ x+log⁡∣x−1x+1∣+cx + \log \left| {\dfrac{{x - 1}}{{x + 1}}} \right| + c ”.

Note : We have to necessarily add an arbitrary constant, let's say c, in the final answer after solving an indefinite integral. Care should be taken while manipulating the numerator and denominator so as to convert the integral into a form that is easier to integrate.