How do you express $ \dfrac{3x-1}{{{x}^{2}}-x} $ in partial fractions?

Hint: In this question, we have to express the given algebraic term in terms of the partial fractions. The partial fraction is the method where we convert the denominator into the product of the two polynomials and then split it in terms of the addition. So, we will apply the partial fractions method to solve the problem. We will first change the denominator into the product of two polynomials and then we will split the numerator into the factors of constant say A and B. After that, we will multiply both sides of the equation by the product of two polynomials. Then, we will calculate the value of constants A and B, which gives the required solution to the problem.

Complete step by step answer:
According to the problem, we have to express an algebraic term into partial fractions.
So, we will simply use the partial fraction method to solve the problem
The algebraic term given to us is $ \dfrac{3x-1}{{{x}^{2}}-x} $ --------- (1)
So, we first rewrite the denominator of equation (1) into the product of two polynomials, that is we take common x from the denominator, we get
 $ {{x}^{2}}-x=x(x-1) $ ----------- (2)
Now, let the numerator of equation (1) equals the sum of two factors, say A and B, that is
 $ 3x-1=A+B $ --------- (3)
Since the factors are linear then the numerators of the partial fractions will be constants say A and B as seen in equation (3), therefore we get

\[\dfrac{3x-1}{x(x-1)}=\dfrac{A}{x}+\dfrac{B}{x-1}\] ----------- (4)
Now, we will take the LCM on the right-hand side of the above equation, we get
\[\dfrac{3x-1}{x(x-1)}=\dfrac{A(x-1)+B.x}{x(x-1)}\]
Now, we will multiply the value of equation (2) in equation (3), we get
$\Rightarrow$ \[3x-1=A(x-1)+Bx\]
Now, we apply the distributive property $ a(b-c)=ab-ac $ in the above equation, we get
\[3x-1=Ax-A+Bx\]
Therefore, we get
$\Rightarrow$ \[3x-1=Ax+Bx-A\]
Now, we will take common x from the above equation, we get
\[3x-1=(A+B)x-A\]
On comparing the left-hand side and the right-hand side of the above equation, we get
 $ 3=A+B $ --------- (5) and
 $ -1=-A $ ---------- (6)
Now, we first solve the equation (6), by multiplying both sides o the equation by (-1), we get
 $ -1.(-1)=-A.(-1) $
Therefore, we get
 $ A=1 $
Now, we will substitute the above value of A in the equation (5), we get
$\Rightarrow$ $ 3=1+B $
Now, we will subtract 1 on both sides of the above equation, we get
$\Rightarrow$ $ 3-1=1+B-1 $
As we know, the same terms with opposite signs cancel out each other, therefore we get
 $ 2=B $
$\Rightarrow$ Thus, now we will put the value of A and B in equation (4), we will get
\[\dfrac{3x-1}{x(x-1)}=\dfrac{1}{x}+\dfrac{2}{x-1}\] which is our required answer.
Therefore, for the equation \[\dfrac{3x-1}{{{x}^{2}}-x}\] , its partial fraction is \[\dfrac{1}{x}+\dfrac{2}{x-1}\].

Note:
 While solving this problem, mention all the formulas and the method you are using to avoid errors and confusion. While splitting the denominator into the product of the two polynomials, do take the common variable and the constant and then further solve, you will get the accurate answer for your problem.