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

Question

Emma Aldridge · Accepted Answer

#int(4x-1)/(x^2(x-4))dx=-15/16lnx-1/(4x)+15/16ln(x-4)# Partial fractions of #(4x-1)/(x^2(x-4))# will be of type #A/x+B/x^2+C/(x-4)# i.e. #(4x-1)/(x^2(x-4))hArrA/x+B/x^2+C/(x-4)# or #hArr(Ax(x-4)+B(x-4)+Cx^2)/(x^2(x-4))# or #hArr(Ax^2-4Ax+Bx-4B+Cx^2)/(x^2(x-4))# or #(4x-1)/(x^2(x-4))hArr((A+C)x^2+x(-4A+B)-4B)/(x^2(x-4))# Hence #A+C=0#, #-4A+B=4# and #4B=1# i.e. #B=1/4#, #-4A=4-1/4=15/4# or #A=-15/16# and #C=15/16# and #(4x-1)/(x^2(x-4))hArr-15/(16x)+1/(4x^2)+15/(16(x-4))# Hence #int(4x-1)/(x^2(x-4))dxhArrint-15/(16x)dx+int1/(4x^2)dx+int15/(16(x-4))dx# = #-15/16lnx-1/(4x)+15/16ln(x-4)#

Ezra Adams · Answer

undefined To integrate $ \frac{4x - 1}{x^2(x - 4)} $ using partial fractions, first factor the denominator. Then express the fraction as the sum of simpler fractions.

$ \frac{4x - 1}{x^2(x - 4)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 4} $

Next, clear the denominators by multiplying both sides by $ x^2(x - 4) $.

$ 4x - 1 = A(x)(x - 4) + B(x - 4) + C(x^2) $

Now, solve for $ A $, $ B $, and $ C $ by equating coefficients of corresponding terms.

Once you have found the values of $ A $, $ B $, and $ C $, you can rewrite the original expression as:

$ \frac{4x - 1}{x^2(x - 4)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 4} $

Then integrate each term separately.