Question

How do you integrate $\dfrac{{{x^2} + 2x - 1}}{{{x^3} - x}}dx?$

Answer 1

Hint: In this question, we need to evaluate the given equation. For this, we will use the partial fraction method and apply the defined integral formulae.

Complete step by step answer:
Let the given integral be $I$ ,so, we can write $I = \int {\dfrac{{{x^2} + 2x - 1}}{{{x^3} - x}}dx} $
Now, by following the partial fraction method, we can write the function $\dfrac{{{x^2} + 2x - 1}}{{{x^3} - x}}$ as
$\dfrac{{{x^2} + 2x - 1}}{{{x^3} - x}} = \dfrac{{{x^2} + 2x - 1}}{{x({x^2} - 1)}}$
We know that the ${1^2} = 1$ , substitute this in the position of $1$ and we get
$ = \dfrac{{{x^2} + 2x - 1}}{{x({x^2} - {1^2})}}$
We use the formula ${a^2} - {b^2} = (a + b)(a - b)$ in the above function and we get
$ = \dfrac{{{x^2} + 2x - 1}}{{x(x + 1)(x - 1)}}$
We use the partial fraction method and we get
$\dfrac{{{x^2} + 2x - 1}}{{{x^3} - x}} = \dfrac{A}{x} + \dfrac{B}{{x - 1}} + \dfrac{C}{{x + 1}}$
$ = \dfrac{{A(x + 1)(x - 1) + Bx(x + 1) + Cx(x - 1)}}{{x(x + 1)(x - 1)}}$
$ = \dfrac{{A({x^2} - 1) + B({x^2} + x) + C({x^2} - x)}}{{{x^3} - x}}$
$ = \dfrac{{(A + B + C){x^2} + (B - C)x - A}}{{{x^3} - x}}$
Therefore $\dfrac{{{x^2} + 2x - 1}}{{{x^3} - x}} = \dfrac{{(A + B + C){x^2} + (B - C)x - A}}{{{x^3} - x}}$
Now, comparing the coefficients of both sides of the above equation, we get
$A + B + C = 1$ …………………………………………(i)
$B - C = 2$ ………………………………………………(ii)
$A = 1$ …………………………………………………(iii)
Solving above three equations and we get
From (iii) we get $A = 1$
Substitute (iii) in (i), we get
$ \Rightarrow 1 + B + C = 1$
Simplifying, we get
$ \Rightarrow B + C = 0$ …………………………………………….(iv)
Adding (ii) and (iv), we get
$(B - C) + (B + C) = 2 + 0$
$ \Rightarrow 2B = 2$
Divide both sides by $2$ , we get
$ \Rightarrow B = 1$
Putting $B = 1$ in equation (iv), we get
$ \Rightarrow 1 + C = 0$
$ \Rightarrow C = - 1$
Hence the values of $A,B,C$ are $1,1, - 1$ respectively.
So , the given integral function can be written as
\[\int {\dfrac{{{x^2} + 2x - 1}}{{{x^3} - x}}} dx = \int {\left( {\dfrac{A}{x} + \dfrac{B}{{x - 1}} + \dfrac{C}{{x + 1}}} \right)} dx\]
\[ = \int {\left( {\dfrac{1}{x} + \dfrac{1}{{x - 1}} + \dfrac{{ - 1}}{{x + 1}}} \right)} dx\]
Now, applying the property of the integration function $\int {\left( {A + B + C} \right)dx = \int {Adx} + \int {Bdx} + \int {Cdx} } $ in the above function, we get
\[ \Rightarrow \int {\dfrac{{{x^2} + 2x - 1}}{{{x^3} - x}}} dx = \int {\left( {\dfrac{1}{x} + \dfrac{1}{{x - 1}} + \dfrac{{ - 1}}{{x + 1}}} \right)} dx\]
\[ = \int {\dfrac{1}{x}dx + \int {\dfrac{1}{{x - 1}}dx} - \int {\dfrac{1}{{x + 1}}dx} } \]
Again, applying the property of the integration function $\int {\dfrac{{dx}}{x} = \log x} $ in the above equation, we get
$ = \log x + \log (x - 1) - \log (x + 1) + c$
Hence, we can see that the value of the integral \[\int {\dfrac{{{x^2} + 2x - 1}}{{{x^3} - x}}} dx\] is $\log x + \log (x - 1) - \log (x + 1) + c$ , where $c$ is the integral constant.

Note:
 It is interesting to note here that for the partial fraction method, if the denominator term is a raised to the power terms then, we need to bifurcate it as $\dfrac{1}{{{{(y + 1)}^2}}} = \dfrac{A}{{y + 1}} + \dfrac{B}{{{{(y + 1)}^2}}}$ where as if the denominator includes a quadratic equation then, we need to bifurcate it as $\dfrac{1}{{(y + 2)({y^2} + 4y + 2)}} = \dfrac{{Ax + B}}{{{y^2} + 4y + 2}} + \dfrac{C}{{y + 2}}$ .