Partial Fractions?

Express the integrands as a sum of partial fraction and evaluate the integrals

#int (x+4)/(x^2+5x-6) dx#

1 Answer
Mar 12, 2018

#int (x+4)/(x^2+5x-6) dx= 2/7ln|x+6| + 5/7 ln |x-1|+C#

Explanation:

Given: #int (x+4)/(x^2+5x-6) dx#

Factor the denominator:

#int (x+4)/(x^2+5x-6) dx= int (x+4)/((x+6)(x-1)) dx#

Setup the partial fractions equation:

#(x+4)/((x+6)(x-1)) = A/(x+6)+ B/(x-1)#

Multiply by the denominator:

#x+4 = A(x-1)+ B(x+6)#

Eliminate B by letting #x = -6#

#-6+4 = A(-6-1)#

#A = -2/-7 = 2/7#

Eliminate A by letting #x = 1#

#1+4 = B(1+6)#

#B = 5/7#

Check:

#2/7 1/(x+6)+ 5/7 1/(x-1) = 2/7 1/(x+6)(x-1)/(x-1)+ 5/7 1/(x-1)(x+6)/(x+6)#

#2/7 1/(x+6)+ 5/7 1/(x-1) = 2/7 (x-1)/((x+6)(x-1))+ 5/7 (x+6)/((x-1)(x+6))#

#2/7 1/(x+6)+ 5/7 1/(x-1) = 1/7 (2x-2)/((x+6)(x-1))+ 1/7 (5x+30)/((x-1)(x+6))#

#2/7 1/(x+6)+ 5/7 1/(x-1) = 1/7 (7x+28)/((x+6)(x-1))#

#2/7 1/(x+6)+ 5/7 1/(x-1) = (x+4)/((x+6)(x-1)) = (x+4)/(x^2+5x-6)#

This checks.

#int (x+4)/(x^2+5x-6) dx= 2/7int 1/(x+6) dx + 5/7 int 1/(x-1) dx#

#int (x+4)/(x^2+5x-6) dx= 2/7ln|x+6| + 5/7 ln |x-1|+C#