How do you use partial fractions to find the integral #int 3/(x^2+x-2) dx#?

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Answer 1 · 2017-01-28T23:28:35

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

Note that #x^2 + x - 2# can be factored as #(x + 2)(x - 1)#.

Thus , our partial fraction decomposition will be of the form

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

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

#Ax- A + Bx + 2B = 3#

#(A + B)x + (2B - A) = 3#

Write a system of equations and solve:

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

Solve either through substitution or elimination to get #A = -1# and #B = 1#.

Therefore, the integral becomes

#int1/(x- 1) - 1/(x + 2)dx#

This can be solved using #int1/xdx = ln|x| + C#.

#=ln|x - 1| - ln|x + 2| + C#

Hopefully this helps!