How do you find the indefinite integral of ∫ ( x^2 - 9x + 14 )dx / (x^2-4x+3) (x-4) ?

Question

Christopher Agresta · Accepted Answer

# ln|(x-1)|+2ln|(x-3)|-2ln|(x-4)|+C, or, ln|{(x-1)(x-3)^2}/(x-4)^2|+C.#

Suppose that, #I=int(x^2-9x+14)/{(x^2-4x+3)(x-4)}dx#.

# :. I=int(x^2-9x+14)/{(x-3)(x-1)(x-4)}dx#.

We will decompose the Integrand #(x^2-9x+14)/{(x-3)(x-1)(x-4)}#

applying the Partial Fraction Method.

To this end, we let, for some #A,B,C in RR#,

# (x^2-9x+14)/{(x-3)(x-1)(x-4)}=A/(x-1)+B/(x-3)+C/(x-4)#.

We use Heaviside's Cover Up Method to determine #A,B,C in RR#.

#:. A=[(x^2-9x+14)/{(x-3)(x-4)}]_(x=1)=(1-9+14)/{(-2)(-3)}=1#,

# B=[(x^2-9x+14)/{(x-1)(x-4)}]_(x=3)=(9-27+14)/{(2)(-1)}=2#,

# C=[(x^2-9x+14)/{(x-3)(x-1)}]_(x=4)=(16-36+14)/{(1)(3)}=-2#.

#:. I=int{1/(x-1)+2/(x-3)-2/(x-4)}dx#,

#=ln|(x-1)|+2ln|(x-3)|-2ln|(x-4)|#.

# rArr I=ln|(x-1)|+2ln|(x-3)|-2ln|(x-4)|+C, or, #

# I=ln|{(x-1)(x-3)^2}/(x-4)^2|+C.#

Take Joy in Mathematics!

Charles Anctil · Answer

undefined To find the indefinite integral of $ \int \frac{x^2 - 9x + 14}{(x^2 - 4x + 3)(x - 4)} \, dx $, you can use partial fraction decomposition. First, factor the denominator into linear factors:

$ x^2 - 4x + 3 = (x - 3)(x - 1) $

Now, rewrite the original integral with the partial fraction decomposition:

$ \int \frac{x^2 - 9x + 14}{(x - 3)(x - 1)(x - 4)} \, dx = \int \left( \frac{A}{x - 3} + \frac{B}{x - 1} + \frac{C}{x - 4} \right) \, dx $

Next, find the constants $ A $, $ B $, and $ C $ by multiplying both sides by the denominator and equating coefficients.

Once you have found the values of $ A $, $ B $, and $ C $, integrate each term separately. The integral of $ \frac{A}{x - 3} $ is $ A \ln|x - 3| $, the integral of $ \frac{B}{x - 1} $ is $ B \ln|x - 1| $, and the integral of $ \frac{C}{x - 4} $ is $ C \ln|x - 4| $.

Combine the results to get the final answer.