How do you integrate #int (1-2x^2)/((x+9)(x+7)(x+1)) # using partial fractions?

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Answer 1 · 2018-02-05T17:24:35

The denominator factors are linear and distinct.

This fact makes the problem a little easier.

#(-2x^2 + 1)/((x + 9)(x + 7)(x + 1)) = A/(x+9) + B/(x + 7) + C/(x + 1)#

Multiply both sides by the denominator on the left side. After cancellation we have:

#-2x^2 + 1 = A(x +7)(x + 1) + B(x + 9)(x + 1) + C(x + 9)(x + 7)#

Short Cut #1:

The above is true for all values of x. In particular it is true for x = -1, which makes one of the factors zero. Let x = -1.

#-2(-1)^2 + 1 = C(-1 + 9)(-1 + 7)#
#-1 = C(8)(6)#
#-1 = 48C#

So #C = -1/48#.

Try the same short cut with x = -7 and then with x = -9, and you will have the values of A, B, and C.

Put those values into #A/(x+9) + B/(x + 7) + C/(x + 1)# and integrate term by term. You will get three natural logarithms if you do it right.