How do you integrate #(3x+2) / (x^(2)+3x-4)dx# using partial fractions?

Question

Emilia Aguilar · Accepted Answer

#ln|x-1|+2ln|x+4|+C#,

or,

#ln|(x-1)(x+4)^2|+C#.

Let #I=int(3x+2)/(x^2+3x-4)dx=int(3x+2)/((x-1)(x+4))dx# To decompose the integrand using Method of Partial Fraction, let, #(3x+2)/((x-1)(x+4))=A/(x-1)+B/(x+4)", where, "A,B in RR#. We use Heavyside's Cover-up Method to determine #A, and, B# : #A=[(3x+2)/(x+4)]_(x=1) =(3+2)/(1+4)=1# #B=[(3x+2)/(x-1)]_(x=-4) =(-12+2)/(-4-1)=2#. Hence, #I=int[1/(x-1)+2/(x+4)]dx# #=int1/(x-1)dx+2int1/(x+4)dx# #=ln|x-1|+2ln|x+4|# Therefore, #I=ln|x-1|+2ln|x+4+C|#, or, #I=ln|(x-1)(x+4)^2|+C#. Enjoy Maths.!

Cameron Alexander · Answer

undefined To integrate $\frac{{3x + 2}}{{x^2 + 3x - 4}}$ using partial fractions, first factor the denominator:

$x^2 + 3x - 4 = (x + 4)(x - 1)$

The partial fraction decomposition will have the form:

$\frac{{3x + 2}}{{x^2 + 3x - 4}} = \frac{{A}}{{x + 4}} + \frac{{B}}{{x - 1}}$

To find $A$ and $B$, multiply both sides by the denominator $x^2 + 3x - 4$ and simplify:

$3x + 2 = A(x - 1) + B(x + 4)$

This gives two equations:

1. $3x + 2 = Ax - A + Bx + 4B$
2. $3x + 2 = (A + B)x + (4B - A)$

By comparing coefficients, we get:

$A + B = 3$
$4B - A = 2$

Solve these simultaneous equations to find $A$ and $B$.

Once you have $A$ and $B$, integrate each term separately:

$\int \frac{{3x + 2}}{{x^2 + 3x - 4}} dx = \int \frac{{A}}{{x + 4}} dx + \int \frac{{B}}{{x - 1}} dx$

Then integrate each term using the natural logarithm function.