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

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Last updated date: 15th Jul 2024
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Answer
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Hint: Analyse the term in the given integral. Use some identity for the numerator of the term and then simplify it. Then use one of the formulas for standard integrals and find an expression for the given integral.
Formula used:
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, where a and b are real numbers.
 \[\int {{x^n}dx} = \left( {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right) + c\] , where c is the integration constant.

Complete step-by-step answer:
Consider the given integral \[I = \int {\left( {\dfrac{{{x^2} - 1}}{{x + 1}}} \right)} dx\] ….. (i)
In the numerator of the integral I we can see the term as \[{x^2} - 1\] .
The term \[{x^2} - 1\] can be written as \[{x^2} - 1 = {x^2} - {1^2}\] because square of 1 is equal to 1.
Now, we can use the identity, which says that ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, where a and b are real numbers.
In this case, we know that x and 1 are real numbers. Therefore, the numerator of the integral I can be written as \[{x^2} - {1^2} = \left( {x + 1} \right)\left( {x - 1} \right)\] .
Substitute this value in equation (i).
 \[ \Rightarrow I = \int {\left[ {\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{{x + 1}}} \right] dx} \] ….. (ii)
Now, we can split the term \[\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{{x + 1}}\] can be written as \[\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{{x + 1}} = x - 1\] .
Substitute the simplified value of \[\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{{x + 1}}\] in equation (ii).
 \[ \Rightarrow I = \int {(x - 1)dx} \]
 \[ \Rightarrow I = \int {xdx - dx} \]
Now, we will be using the distributive property of integration. And by using this property we get that the integral I can be written as
 \[I = \int {xdx} - \int {dx} \] …. (iii)
.We know that \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\] , where c is the integration constant.
Consider, \[\int {xdx} \] .
Here, $n = 1$.
Therefore, \[\int {xdx} = \dfrac{{{{\left( x \right)}^{1 + 1}}}}{{1 + 1}} + c = \dfrac{{{x^2}}}{2} + c\] .
Now, consider \[\int {dx} \]
Here, $n = 0$
Therefore, \[\int {dx} = \dfrac{{{{\left( x \right)}^{0 + 1}}}}{{0 + 1}} + c = x + c\] .
Substitute the known values in equation (iii).
  \[ \Rightarrow I = \dfrac{{{x^2}}}{2} - x + c\] .
Therefore, the given integral \[\int {\dfrac{{{x^2} - 1}}{{x + 1}}dx} \] is equal to \[\dfrac{{{x^2}}}{2} - x + c\] .
So, the correct answer is “ \[\dfrac{{{x^2}}}{2} - x + c\] ”.

Note: It is necessary to put the integration constant after we solve the given integral. This is because the simplified expression for the integral consists of an infinite number of equations for the association differential equation.
One may also check whether the solved integral is correct or not by differentiating the solved expression with respect to the variable and check whether it results to the initial expression of the integral.