How do you integrate #(5x^2-9x)/((x-4)(x-1)^2)# using partial fractions?

Question

Isaiah Allen · Accepted Answer

#int (5x^2-9x)/((x-4)(x-1)^2) dx = 44/9 ln abs(x-4) - 4/(3(x-1)) + 1/9 ln abs(x-1) + C#

#(5x^2-9x)/((x-4)(x-1)^2) = A/(x-4)+B/(x-1)^2+C/(x-1)#

#color(white)((5x^2-9x)/((x-4)(x-1)^2)) = (A(x-1)^2+B(x-4)+C(x-4)(x-1))/((x-4)(x-1)^2)#

#color(white)((5x^2-9x)/((x-4)(x-1)^2)) = (A(x^2-2x+1)+B(x-4)+C(x^2-5x+4))/((x-4)(x-1)^2)#

#color(white)((5x^2-9x)/((x-4)(x-1)^2)) = ((A+C)x^2+(-2A+B-5C)x+(A-4B+4C))/((x-4)(x-1)^2)#

Equating coefficients we get this system of linear equations:

#{ (A+C=5), (-2A+B-5C=-9), (A-4B+4C=0) :}#

Adding all three equations, we find:

#-3B = -4#

So #color(blue)(B=4/3)#

Adding twice the first equation to the second, we get:

#B-3C=1#

Hence:

#3C = B-1 = 4/3-1 = 1/3#

So #color(blue)(C=1/9)#

Then from the first equation:

#color(blue)(A=)5-C = 5-1/9 = color(blue)(44/9)#

So:

#int (5x^2-9x)/((x-4)(x-1)^2) dx = int (44/9 * 1/(x-4)+4/3 * 1/(x-1)^2 +1/9 * 1/(x-1)) dx#

#color(white)(int (5x^2-9x)/((x-4)(x-1)^2) dx) = 44/9 ln abs(x-4) - 4/(3(x-1)) + 1/9 ln abs(x-1) + #C

Paisley Johnson · Answer

undefined To integrate $(5x^2-9x)/((x-4)(x-1)^2)$ using partial fractions, follow these steps:

1. Perform long division if necessary to ensure that the degree of the numerator is less than the degree of the denominator.
2. Write the expression in the form $\frac{A}{x-4} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$.
3. Multiply both sides by the denominator $(x-4)(x-1)^2$ to clear the fractions.
4. Solve for the constants $A$, $B$, and $C$ by comparing coefficients.
5. Integrate each term separately.
6. Combine the results to get the final integral.

The result after integration will be:

$\frac{5}{3}\ln|x-1| - \frac{3}{x-1} - 4\ln|x-4| + C$, where $C$ is the constant of integration.