How do you integrate #int (x-2x^2)/((x-6)(x-2)(x-5)) # using partial fractions?

Question

Luke Alvarez · Accepted Answer

Divide by #x-2# first

#x/[(x-6)(x-5)] = x[1/(x-6) - 1/(x-5)]# #= 6/(x-6) - 5/(x-5)# And the integral I is #6 log |x-6| - 5 log|x-5|#

Samantha Adkison · Answer

undefined To integrate the given rational function using partial fractions, first factor the denominator into irreducible quadratic factors:

(x - 6)(x - 2)(x - 5)

Now, express the integrand as the sum of partial fractions:

(x - 2x^2)/((x - 6)(x - 2)(x - 5)) = A/(x - 6) + B/(x - 2) + C/(x - 5)

Next, clear the denominators by multiplying both sides by the denominator:

x - 2x^2 = A(x - 2)(x - 5) + B(x - 6)(x - 5) + C(x - 6)(x - 2)

Expand and combine like terms:

x - 2x^2 = A(x^2 - 7x + 10) + B(x^2 - 11x + 30) + C(x^2 - 8x + 12)

Now, equate the coefficients of like terms on both sides:

For x^2 term: -2 = A + B + C
For x term: 1 = -7A - 11B - 8C
For constant term: 0 = 10A + 30B + 12C

Now, solve this system of linear equations to find the values of A, B, and C. Once you have found A, B, and C, substitute them back into the partial fraction decomposition:

A/(x - 6) + B/(x - 2) + C/(x - 5)

Then, integrate each term individually. The resulting integrals should be straightforward to solve.