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

Question

Elijah Abram · Accepted Answer

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!

Emilia Aguilar · Answer

undefined To integrate the function $ \frac{3}{x^2 + x - 2} $ using partial fractions, first factor the denominator as $ (x + 2)(x - 1) $. Then, express $ \frac{3}{x^2 + x - 2} $ as the sum of two fractions with undetermined coefficients, $ \frac{A}{x + 2} $ and $ \frac{B}{x - 1} $.

Solve for $ A $ and $ B $ by equating the original expression to the partial fraction decomposition. This leads to $ A = 1 $ and $ B = 2 $.

Therefore, the integral becomes $ \int \frac{3}{x^2 + x - 2} dx = \int \frac{1}{x + 2} dx + \int \frac{2}{x - 1} dx $.

Now, integrate each term separately to get $ \ln|x + 2| + 2\ln|x - 1| + C $, where $ C $ is the constant of integration.