How do you evaluate #\int \frac { d x } { ( x + 2) ( x ^ { 2} - 1) }#?

1 Answer
Noah G
Apr 2, 2017

The integral equals #1/3ln|x + 2| - 1/2ln|x + 1| + 1/6ln|x - 1| + C#

Explanation:

Note that #(x + 2)(x^2 - 1)# can be written as #(x + 2)(x + 1)(x - 1)#. By partial fractions, we have:

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

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

#Ax^2 - A + B(x^2 + x - 2) + C(x^2 + 3x + 2) = 1#

#Ax^2 - A + Bx^2 + Bx - 2B + Cx^2 + 3Cx + 2C = 1#

#(A + B + C)x^2 + (B + 3C)x + (2C - 2B - A) = 1#

Write a system of equations now.

#{(A + B + C = 0), (B + 3C = 0), (2C - 2B - A = 1):}#

We solve.

#B = -3C -> 2C - 2(-3C) - A = 1#

Also, #A = -B - C -> A = -(-3C) - C#

#2C + 6C - (3C - C) = 1#

#8C - 3C + C = 1#

#C = 1/6#

Solving for the other variables we get #B = -1/2# and #A = 1/3#.

The integral becomes

#int 1/(3(x + 2)) - 1/(2(x + 1)) + 1/(6(x - 1))dx#

Which can be integrated as

#1/3ln|x + 2| - 1/2ln|x + 1| + 1/6ln|x - 1| + C#

Hopefully this helps!

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