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!

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