We wish to compute integral_-3^3 x^2 + 3/x^3 + 4x^2 - 16x - 64 dx. (i) We begin

Question

We wish to compute integral_-3^3 x^2 + 3/x^3 + 4x^2 - 16x - 64 dx. (i) We begin by factorizing the denominator of the rational function. We get x^3 + 4x^2 - 16x - 64 = (x- a) (x - b)^2 for a notequalto b. What are a and b? a = , b = (ii) Next, we express the fraction in the form x^2 + 3/x^3 + 4x^2 - 16x - 64 = A/x - a + B/x - b + C/(x - b)^2. Give the exact values of A , B and C. A = , B = , C = (iii) Finally, we use this partial fraction decomposition to compute the integral. Give its approximate value with 3 decimal places. integral_-3^3 x^ + 3/x^3 + 4x^2 - 16x - 64 dx approximately

Explanation / Answer

we have given integration of (x=-3 to 3) (x2+3)/(x3+4x2-16x-64)

(i) factoring the denominator

(x3+4x2-16x-64) =(x-4)(x-(-4))2

a=4,b=-4

ii) expressing the partial fraction form

(x2+3)/[(x-4)(x-(-4))2] =A/x-4 +B/(x-(-4)) +C/(x-(-4))2

=[A(x-(-4))^2+B(x-4)(x-(-4))+C(x-4)]/[(x-4)(x-(-4))^2]

multiply the equation by denominator on both sides

[(x2+3)*(x-4)(x-(-4))2]/[(x-4)(x-(-4))2] =[A(x-(-4))^2+B(x-4)(x-(-4))+C(x-4)]*((x-4)(x-(-4))^2)/[(x-4)(x-(-4))^2]

(x2+3)=[A(x-(-4))^2+B(x-4)(x-(-4))+C(x-4)]

plug the real roots of denominator x=,4,-4

plug x=4 into above equation

(16+3)=64A

A=19/64

plug x=-4 into above equation

-8C=19

C=-19/8

plug the A=19/64,C=-19/8 into above equation

(x2+3)=[(19/64)(x-(-4))^2+B(x-4)(x-(-4))-(19/8)(x-4)]

(x2+3)=(19/64)x^2+(19/64)*16+(19/64)*8x+Bx^2-B16-(19/8)x+(19/8)*4

group elements according to powers of x

(x2+3)=((19/64)+B)x^2+(-B16+57/4)

-B16+57/4=3

16B=57/4-3=45/4

B=45/4/16 =45/64

(x2+3)/[(x-4)(x-(-4))2] =19/64(x-4) + 45/(64(x-(-4))) -19/(8(x-(-4))2)

A=19/64,B=45/64,C=-19/8

integration of (x=-3 to 3) (x2+3)/(x3+4x2-16x-64)

=[integration of (19/64(x-4) + 45/(64(x+4)) -19/(8(x+4)2))]dx from x=-3 to 3 ---(1)

now solving integration of (19/(64(x-4)))dx

substitute u=x-4,dx=du

integration of (19/(64(x-4)))dx= integration of (19/(64(u)))du =(19/64)*ln(u)=(19/64)*ln|(x-4)| since u=x-4

now solving integration of (45/(64(x+4)))dx

substitute u=x+4,du=dx

integration of (45/(64(x+4)))dx=integration of (45/(64(u)))du =(45/46)*ln(u)=(45/46)*ln|(x+4)| since u=x+4

now solving integration of (19/(8(x+4)2))dx

substitute u=x+4, du=dx

integration of (19/(8(x+4)2))dx=(19/8)*integration of (u-2)du =-(19/8)*(1/u)=-(19/8)*(1/(x+4)) since u=x+4

substitute above integration values into equation (1)

integration of (x=-3 to 3) (x2+3)/(x3+4x2-16x-64)

=[integration of (19/64(x-4) + 45/(64(x+4)) -19/(8(x+4)2))]dx from x=-3 to 3 ---(1)

=[(19/64)*ln|(x-4)|+(45/46)*ln|(x+4)|+(19/8)*(1/(x+4))] from x=-3 to 3

=[(19/64)*ln|(-1)|+(45/46)*ln|(7)|+(19/8)*(1/7))]-[(19/64)*ln|(-7)|+(45/46)*ln|(1)|+(19/8)*(1/1)]

=(45/64)*ln(7)+19/56-(19/64)*ln(7)-19/8

=(45/64)*ln(7)+(19/56)-(19/64)*(8+ln(7))

=-1.245

integration of (x=-3 to 3) (x2+3)/(x3+4x2-16x-64) =-1.245

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