What is #int (4-x ) / (x^3-6x +4 )#?

Question

What is #int (4-x     ) / (x^3-6x  +4     )#?

Emilia Aguilar · Accepted Answer

#int (4-x)/(x^3-6x+4) dx =#

#=1/3 ln abs(x-2) + (-1-2sqrt(3))/6 ln abs(x+1-sqrt(3))+ (-1+2sqrt(3))/6 ln abs(x+1+sqrt(3)) + C# Express as a partial fraction decomposition first: #x^3-6x+4# #= (x-2)(x^2+2x-2)# #= (x-2)(x^2+2x+1-3)# #=(x-2)(x+1-sqrt(3))(x+1+sqrt(3))# Solve: #(4-x)/(x^3-6x+4)# #=A/(x-2) + B/(x+1-sqrt(3)) + C/(x+1+sqrt(3))# #=(A(x^2+2x-2)+B(x-2)(x+1+sqrt(3))+C(x-2)(x+1-sqrt(3)))/(x^3-6x+4)# #=(A(x^2+2x-2)+B(x^2-(1-sqrt(3))x-2(1+sqrt(3)))+C(x^2-(1+sqrt(3))x-2(1-sqrt(3))))/(x^3-6x+4)# #=((A+B+C)x^2+(2A-(1-sqrt(3))B-(1+sqrt(3))C)x-2(A+(1+sqrt(3))B+(1-sqrt(3))C))/(x^3-6x+4)# Equating coefficients, we get the following simulataneous equations: #(a)color(white)(-)A+B+C = 0# #(b)color(white)(-)2A-(1-sqrt(3))B-(1+sqrt(3))C = -1# #(c)color(white)(-)A+(1+sqrt(3))B+(1-sqrt(3))C = -2# Subtracting #(c)# from #(b)#, we get: #(d)color(white)(-)A-2B-2C = 1# Adding twice #(a)# the first equation to #(d)#, we get: #3A = 1# So #A = 1/3# Adding #(b)+(c)# we get: #(e)color(white)(-)3A+2sqrt(3)B-2sqrt(3)C=-3# We know #3A = 1#, so this simplifies to: #2sqrt(3)B-2sqrt(3)C=-4# Hence: #(f)color(white)(-)B-C = -2/sqrt(3) = -2sqrt(3)/3# From #(a)# we get: #(g)color(white)(-)B+C = -1/3# Then #(f)+(g)# gives us: #2B = (-1-2sqrt(3))/3# and #(g)-(f)# gives us: #2C = (-1+2sqrt(3))/3# Hence: #(4-x)/(x^3-6x+4)# #=1/(3(x-2)) + (-1-2sqrt(3))/(6(x+1-sqrt(3))) + (-1+2sqrt(3))/(6(x+1+sqrt(3)))# Then use: #int 1/t dt = ln abs(t) + C# to find: #int (4-x)/(x^3-6x+4) dx =# #int 1/(3(x-2)) + (-1-2sqrt(3))/(6(x+1-sqrt(3))) + (-1+2sqrt(3))/(6(x+1+sqrt(3))) dx# #=1/3 ln abs(x-2) + (-1-2sqrt(3))/6 ln abs(x+1-sqrt(3))+ (-1+2sqrt(3))/6 ln abs(x+1+sqrt(3)) + C#

Cameron Alexander · Answer

undefined The integral of $ \frac{4-x}{x^3-6x+4} $ is calculated using partial fraction decomposition followed by integrating each term. The steps involve:

1. Factorizing the denominator.
2. Performing partial fraction decomposition.
3. Integrating each term separately.