What is int (4-x ) / (x^3-6x +4 )∫4−xx3−6x+4?

1 Answer
George C.
Mar 15, 2016

int (4-x)/(x^3-6x+4) dx =∫4−xx3−6x+4dx=

=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=13ln|x−2|+−1−2√36ln∣∣x+1−√3∣∣+−1+2√36ln∣∣x+1+√3∣∣+C

Explanation:

Express as a partial fraction decomposition first:

x^3-6x+4x3−6x+4

= (x-2)(x^2+2x-2)=(x−2)(x2+2x−2)

= (x-2)(x^2+2x+1-3)=(x−2)(x2+2x+1−3)

=(x-2)(x+1-sqrt(3))(x+1+sqrt(3))=(x−2)(x+1−√3)(x+1+√3)

Solve:

(4-x)/(x^3-6x+4)4−xx3−6x+4

=A/(x-2) + B/(x+1-sqrt(3)) + C/(x+1+sqrt(3))=Ax−2+Bx+1−√3+Cx+1+√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(x2+2x−2)+B(x−2)(x+1+√3)+C(x−2)(x+1−√3)x3−6x+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(x2+2x−2)+B(x2−(1−√3)x−2(1+√3))+C(x2−(1+√3)x−2(1−√3))x3−6x+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)=(A+B+C)x2+(2A−(1−√3)B−(1+√3)C)x−2(A+(1+√3)B+(1−√3)C)x3−6x+4

Equating coefficients, we get the following simulataneous equations:

(a)color(white)(-)A+B+C = 0(a)−A+B+C=0

(b)color(white)(-)2A-(1-sqrt(3))B-(1+sqrt(3))C = -1(b)−2A−(1−√3)B−(1+√3)C=−1

(c)color(white)(-)A+(1+sqrt(3))B+(1-sqrt(3))C = -2(c)−A+(1+√3)B+(1−√3)C=−2

Subtracting (c)(c) from (b)(b), we get:

(d)color(white)(-)A-2B-2C = 1(d)−A−2B−2C=1

Adding twice (a)(a) the first equation to (d)(d), we get:

3A = 13A=1

So A = 1/3A=13

Adding (b)+(c)(b)+(c) we get:

(e)color(white)(-)3A+2sqrt(3)B-2sqrt(3)C=-3(e)−3A+2√3B−2√3C=−3

We know 3A = 13A=1, so this simplifies to:

2sqrt(3)B-2sqrt(3)C=-42√3B−2√3C=−4

Hence:

(f)color(white)(-)B-C = -2/sqrt(3) = -2sqrt(3)/3(f)−B−C=−2√3=−2√33

From (a)(a) we get:

(g)color(white)(-)B+C = -1/3(g)−B+C=−13

Then (f)+(g)(f)+(g) gives us:

2B = (-1-2sqrt(3))/32B=−1−2√33

and (g)-(f)(g)−(f) gives us:

2C = (-1+2sqrt(3))/32C=−1+2√33

Hence:

(4-x)/(x^3-6x+4)4−xx3−6x+4

=1/(3(x-2)) + (-1-2sqrt(3))/(6(x+1-sqrt(3))) + (-1+2sqrt(3))/(6(x+1+sqrt(3)))=13(x−2)+−1−2√36(x+1−√3)+−1+2√36(x+1+√3)

Then use: int 1/t dt = ln abs(t) + C∫1tdt=ln|t|+C to find:

int (4-x)/(x^3-6x+4) dx =∫4−xx3−6x+4dx=

int 1/(3(x-2)) + (-1-2sqrt(3))/(6(x+1-sqrt(3))) + (-1+2sqrt(3))/(6(x+1+sqrt(3))) dx∫13(x−2)+−1−2√36(x+1−√3)+−1+2√36(x+1+√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=13ln|x−2|+−1−2√36ln∣∣x+1−√3∣∣+−1+2√36ln∣∣x+1+√3∣∣+C

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