How do you integrate #int (2x - 1) / (x^2 + 5x - 6) dx# using partial fractions?

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Answer 1 · 2016-10-31T13:49:12

The integral equals # 13/7ln|x + 6| + 1/7ln|x - 1| + C#.

We can factor the denominator as #(x + 6)(x - 1)#.

So,

#A/(x + 6) + B/(x- 1) = (2x - 1)/((x + 6)(x - 1))#

#A(x - 1) + B(x + 6) = 2x - 1#

#Ax - A + Bx + 6B = 2x - 1#

#(A + B)x +(-A + 6B) = 2x - 1#

We can now write a systems of equations to solve for #A# and #B#.

#A + B = 2#
#-A + 6B = -1#

#B = 2- a#

#-A + 6(2 - A) = -1#

#-A + 12 - 6A = -1#

#-7A = -13#

#A = 13/7#

#13/7 + B = 2#

#B = 2- 13/7#

#B = 1/7#

Resubstitute into the initial equation.

The integral become #int((13/7)/(x + 6) + (1/7)/(x - 1)) dx#

We now use the rule #int(1/x)(dx) = ln|x| + C# to get rid of the integral.

Letting the function be #f(x)#.

#f(x) = 13/7ln|x + 6| + 1/7ln|x - 1| + C#

Hopefully this helps!