Evaluate the definite integral.?

Question

integral from [0,1] of (x^3-1)/(x^2-4)dx

I realize that this is an improper rational function where you have to divide the top by the bottom and after I solved for the entire integral, I got an answer that was totally wrong. I understand you set up the equation of A/(x+2) + B/(x-2) = 4x-1. Can someone please help me?

Answer 1

a / [ x + 2 ] + b / [ x - 2 ] = { 4x - 1 } / [ x² - 4 ]...



multiply both sides by x + 2 and let x = - 2 to find a , by x - 2 and let x = 2 to find b



{ 9 / 4 , 7 / 4 } .. ...answer x² + [ 9 / 4 ] ln [ x + 2 ] + [ 7 / 4 ] ln [ x - 2 ] + C

Answer 2

Hello,



₁

∫ [(x³ - 1) /(x² - 4)] dx =

⁰



let's figure out the antiderivative, dividing the numerator by the denominator:



..x³. + 0x². + 0x. - 1 | x² - 4

- x³...........+ 4x.......| x

------------------------------

..0............. .4x. - 1



yielding:



∫ {x + [(4x - 1) /(x² - 4)]} dx =



(splitting into two integrals)



∫ x dx + ∫ [(4x - 1) /(x² - 4)] dx =



[1/(1+1)] x^(1+1) + ∫ [(4x - 1) /(x² - 4)] dx =



(1/2)x² + ∫ [(4x - 1) /(x² - 4)] dx (#)



let's factor the denominator and decompose the remaining integrand into partial fractions:



(4x - 1) /(x² - 4) = (4x - 1) /[(x + 2)(x - 2)] = A/(x + 2) + B/(x - 2)



(4x - 1) /[(x + 2)(x - 2)] = [A(x - 2) + B(x + 2)] /[(x + 2)(x - 2)]



(equating numerators)



4x - 1 = Ax - 2A + Bx + 2B



4x - 1 = (A + B)x + (- 2A + 2B)



(equating coefficients)



A + B = 4

- 2A + 2B = - 1



A = - B + 4

- 2(- B + 4) + 2B = - 1 → 2B - 8 + 2B = - 1 → 4B = - 1 + 8 → 4B = 7



A = - B + 4 = - (7/4) + 4 = (- 7 + 16)/4 = 9/4

B = 7/4



yielding:



(4x - 1) /(x² - 4) = (4x - 1) /[(x + 2)(x - 2)] = A/(x + 2) + B/(x - 2) = (9/4)/(x + 2) + (7/4)/(x - 2)



thus the above (#) expression becomes:



(1/2)x² + ∫ {[(9/4) /(x + 2)] + [(7/4) /(x - 2)]} dx =



(splitting into two integrals and pulling constants out)



(1/2)x² + (9/4) ∫ [1 /(x + 2)] dx + (7/4) ∫ [1 /(x - 2)] dx =



(1/2)x² + (9/4) ln |x + 2| + (7/4) ln |x - 2| + C (antiderivative)



we have the antiderivative, then let's plug in the bounds:



₁

∫ [(x³ - 1) /(x² - 4)] dx = [(1/2)1² + (9/4) ln |1 + 2| + (7/4) ln |1 - 2|] - [(1/2)0² +

⁰

(9/4) ln |0 + 2| + (7/4) ln |0 - 2|] =



(1/2)1 + (9/4) ln 3 + (7/4) ln |- 1| - (1/2)0 - (9/4) ln 2 - (7/4) ln |- 2| =



(1/2) + (9/4) ln 3 + (7/4) ln 1 - 0 - (9/4) ln 2 - (7/4) ln 2 =



(1/2) + (9/4) ln 3 + (7/4)0 - (9/4) ln 2 - (7/4) ln 2 =



(1/2) + (9/4) (ln 3 - ln 2) + 0 - (7/4) ln 2 =



(applying logarithm properties)



(1/2) + (9/4) ln (3/2) - (7/4) ln 2





the result is:



(1/2) + (9/4) ln (3/2) - (7/4) ln 2 (≈ 0.19928)





I hope it's helpful

Answer 3

(x^3-1)/(x^2 - 4) = (x^3 - 1)/((x - 2)(x + 2))



(x^3-1)/(x^2 - 4) = x + A/(x - 2) + B/(x + 2)



(x^3-1)/(x^2 - 4) = (x(x^2 - 4) + A(x + 2)+ B(x - 2))/(x^2 - 4)



check for numerator only,



for x = -2



B(-2 - 2) = (-2)^3 - 1



B = 9/4





for x = 2



A(2 + 2) = (2)^3 - 1



A = 7/4







(x^3-1)/(x^2 - 4) = x + 7/(4(x - 2)) + 9/(4(x + 2))



. . . .1

k = ∫ (x^3-1)/(x^2 - 4) dx

. . . 0



. . . .1

k = ∫ x + 7/(4(x - 2)) + 9/(4(x + 2)) dx

. . . 0



. . . .1. . . . . . . . . . . . . . . 1. . . . . . . . . . . . . . . . .1

k = ∫ 7/(4(x - 2)) d(x - 2) + ∫ 9/(4(x + 2)) d(x + 2) + ∫ x dx

. . . .0. . . . . . . . . . . . . . . 0. . . . . . . . . . . . . . . . .0



. . . ... . . . . . . . . . . . . . . . . . . . . . . . . 1

k = ½ x² + 7/4 ln |x - 2| + 9/4 ln |x + 2| . .|

. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . .0



k = ½ * 1² + 7/4 ln |1 - 2| + 9/4 ln |1 + 2| - (½ * 0² + 7/4 ln |0 - 2| + 9/4 ln |0 + 2|)



k = 0.199289

Answer 4

hi, i comprehend that quite a few persons are having difficulty with the addition sign displaying up on their questions. are you able to make clean the 1st area, is it: (5x^2 + 4)? From there you ought to use integration via areas. i'm rather rusty yet i will have a go at it. :o)

Answer 5

f(x)=x+9/(4*(x+2))+7/(4*(x-2))



I = int(f(x)) dx = (1/2)*x^2+(9/4)*ln(x+2)+(7/4)*ln(x-2) + constant



then from from [0,1]:



I = -4*ln(2)+1/2+(9/4)*ln(3)