What is the integral of #(x+1)/(x(x^2+x-6))#?

Question

Emilia Aguilar · Accepted Answer

#-1/6ln|x| - 2/15 ln|x+3| + 3/10 ln|x-2| + C# Well, for one, we can factor the denominator: #int(x+1)/(x(x^2 + x - 6))dx# #int(x + 1)/(x(x+3)(x-2))dx -= int A/(x) + B/(x+3) + C/(x-2)dx# Now we just have linear denominators, which is a fairly straightforward decomposition.  By achieving common denominators on the righthand side, we can equate the numerator to #x+1#. Therefore, multiply the top and bottom so that all the terms have the denominator #x(x^2 + x - 6)#. #[A(x+3)(x-2) + B(x)(x-2) + C(x)(x+3)]/cancel(x(x+3)(x-2)) = (x+1)/cancel(x(x+3)(x-2))# Simplify the left side so that it looks like a general polynomial. First, distribute: #Ax^2 + Ax - 6A + Bx^2 - 2Bx + Cx^2 + 3Cx = x + 1# Group together terms. #Ax^2 + Bx^2 + Cx^2 + Ax - 2Bx + 3Cx - 6A = x + 1# Now make sure you get it into this correct form, #bb(ax^2 + bx + c)#: #ul((A + B + C))x^2 + ul((A - 2B + 3C))x + ul((-6A))# #= ul(0)x^2 + ul(1)x + ul(1)# This means we have a system of three equations: #A + B + C = 0#
#A - 2B + 3C = 1#
#-6A = 1# Clearly, #color(green)(A = -1/6)#, so the rest follows. Add the negative of the second equation to the first. #" "A + B + C = 0#
#- (A - 2B + 3C = 1)#
#"------------------------------"#
#" "" "" "3B - 2C = -1# This gives #C = (-1 - 3B)/(-2) = 1/2 + 3/2B#, so from the first equation: #-1/6 + B + 1/2 + 3/2B = 0# #1/6 - 1/2 = 5/2B# #=> color(green)(B = -2/15)# Thus: #color(green)(C) = 1/2 + 3/2*-2/15 = color(green)(3/10)# You can verify #A#, #B#, and #C# by plugging them back into the system of equations. And so, we have: #int(x + 1)/(x^3 + x^2 - 6x)dx# #= -1/6int 1/(x)dx - 2/15 int 1/(x+3)dx + 3/10 int 1/(x-2)dx# We know the integral of #1/u# to be #ln|u|#. Thus: #=> color(blue)(int(x + 1)/(x^3 + x^2 - 6x)dx)# #color(blue)(= -1/6ln|x| - 2/15 ln|x+3| + 3/10 ln|x-2| + C)#

Evelyn Agard · Answer

undefined To find the integral of $(x+1)/(x(x^2+x-6))$, you first factor the denominator, then use partial fraction decomposition. Factoring the denominator gives $x(x^2+x-6) = x(x+3)(x-2)$. The partial fraction decomposition of $(x+1)/(x(x^2+x-6))$ is $A/x + B/(x+3) + C/(x-2)$, where $A$, $B$, and $C$ are constants. Solving for these constants yields $A = 1/2$, $B = -1/4$, and $C = 1/4$. Therefore, the integral becomes $(1/2)\ln|x| - (1/4)\ln|x+3| + (1/4)\ln|x-2| + C$, where $C$ is the constant of integration.