How do you integrate #int (4x-2) /( 3(x-1)^2)# using partial fractions?

Question

How do you integrate #int  (4x-2) /( 3(x-1)^2)# using partial fractions?

Scarlett Allison · Accepted Answer

#int(4x-2)/(3(x-1)^2)dx=4/3ln(x-1)-4/(3(x-1))+c# Let us first find partial fractions of #(4x-2)/(3(x-1)^2)# and for this let #(4x-2)/((x-1)^2)hArrA/(x-1)+B/(x-1)^2# or #(4x-2)/((x-1)^2)hArr(A(x-1)+B)/((x-1)^2)=(Ax+B-A)/((x-1)^2)# Hence #A=4# and #B-A=-2# i.e. #B=4-2=2# Hence #int(4x-2)/(3(x-1)^2)dx=1/3int[2/(x-1)+2/(x-1)^2]dx# = #2/3int2/(x-1)dx+2/3int2/(x-1)^2dx+k# = #4/3ln(x-1)-4/(3(x-1))+c#

Christopher Agresta · Answer

undefined To integrate $\frac{{4x - 2}}{{3(x - 1)^2}}$ using partial fractions, you first express the rational function as the sum of two or more simpler fractions.

1. Factor the denominator: $3(x - 1)^2$ factors into $3(x - 1)(x - 1)$.
2. Write the partial fraction decomposition: $\frac{{4x - 2}}{{3(x - 1)^2}} = \frac{{A}}{{3(x - 1)}} + \frac{{B}}{{3(x - 1)^2}}$.
3. Clear the denominators by multiplying both sides of the equation by $3(x - 1)^2$.
4. Solve for $A$ and $B$ by comparing coefficients.
5. Integrate each partial fraction separately.
6. Combine the results to find the final integrated expression.

After solving for $A$ and $B$, you should get:

$$A = 2 \quad \text{and} \quad B = \frac{2}{3}$$

The integral of $\frac{{4x - 2}}{{3(x - 1)^2}}$ using partial fractions becomes:

$$\int{\frac{{4x - 2}}{{3(x - 1)^2}}dx} = \int{\frac{{2}}{{3(x - 1)}}dx} + \int{\frac{{\frac{2}{3}}}{{3(x - 1)^2}}dx}$$

$$= \frac{2}{3}\ln{|x - 1|} - \frac{2}{9(x - 1)} + C$$

Where $C$ is the constant of integration.