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

1 Answer
Mar 15, 2016

#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#

Explanation:

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#