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

Question

Evelyn Agard · Accepted Answer

I decomposed integrand into basic fractions,

#(2x-1)/((x-1)^3*(x-2))=A/(x-1)+B/(x-1)^2+C/(x-1)^3+D/(x-2)#

#(2x-1)=A*(x^3-4x^2+5x-2)+B*(x^2-3x+2)+C*(x-2)+D*(x^3-3x^2+3x-1)#

Let #x=1#, #-C=1# or #C=-1#.

Let #x=2#, #D=3#.

Differentiate both sides,

#2=A*(3x^2-8x+5)+B*(2x-3)+C+D*(3x^2-6x+3)#

Let #x=1#, #-B+C=2# or #-B-1=2#. Hence #B=-3#

Differentiate both sides,

#0,=A*(6x-8)+2B+D*(6x-6)#

Let #x=1#, #-2A+2B=0# or #A=B=-3#.

Thus,

#int(2x-1)/((x-1)^3*(x-2))dx#

#=int-3/(x-1)dx+int-3/(x-1)^2dx++int-1/(x-1)^3dx+int3/(x-2)dx#

#=-3Ln(x-1)+3/(x-1)+1/2*(x-1)^(-2)+3Ln(x-2)+C#

#=3/(x-1)+1/2*(x-1)^(-2)+3Ln((x-2)/(x-1))+C# I broke down the integrand into fundamental fractions.

Emilia Aguilar · Answer

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

Start by factoring the denominator $ (x - 1)^3 (x - 2) $. It factors into linear and quadratic factors as $ (x - 1)(x - 1)(x - 1)(x - 2) $.

Then, set up the partial fraction decomposition as:

$$ \frac{2x - 1}{(x - 1)^3 (x - 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{x - 2} $$

Now, clear the denominators by multiplying both sides by $ (x - 1)^3 (x - 2) $ to get:

$$ 2x - 1 = A(x - 1)^2(x - 2) + B(x - 1)(x - 2) + C(x - 2) + D(x - 1)^3 $$

Next, equate coefficients or use suitable values of $ x $ to solve for $ A $, $ B $, $ C $, and $ D $.

After finding the values of $ A $, $ B $, $ C $, and $ D $, rewrite the original integral as the sum of the integrals of the partial fractions:

$$ \int \frac{2x - 1}{(x - 1)^3 (x - 2)} \, dx = \int \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{x - 2} \, dx $$

Finally, integrate each term individually.

Ezekiel Alexander · Answer

undefined To integrate the expression $\frac{2x - 1}{(x - 1)^3 (x - 2)}$ using partial fractions, we first express the rational function as the sum of simpler fractions:

$$\frac{2x - 1}{(x - 1)^3 (x - 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{x - 2}$$

Next, we find the values of $A$, $B$, $C$, and $D$ by multiplying both sides of the equation by the common denominator $(x - 1)^3 (x - 2)$ and equating coefficients of like terms.

$$2x - 1 = A(x - 1)^2(x - 2) + B(x - 1)(x - 2) + C(x - 2) + D(x - 1)^3$$

By substituting appropriate values for $x$ and solving the resulting system of equations, we can determine the values of $A$, $B$, $C$, and $D$.

Once we have found the values of $A$, $B$, $C$, and $D$, we substitute them back into the partial fraction decomposition:

$$\frac{2x - 1}{(x - 1)^3 (x - 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{x - 2}$$

Now, we can integrate each term individually:

$$\int{\frac{2x - 1}{(x - 1)^3 (x - 2)}dx} = \int{\frac{A}{x - 1}dx} + \int{\frac{B}{(x - 1)^2}dx} + \int{\frac{C}{(x - 1)^3}dx} + \int{\frac{D}{x - 2}dx}$$

Finally, we integrate each term separately using standard integration techniques.