Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Mathematics (https://www.askmehelpdesk.com/forumdisplay.php?f=199)
-   -   Can anyone evaluate these? (https://www.askmehelpdesk.com/showthread.php?t=510775)

  • Sep 25, 2010, 11:01 AM
    gashan
    can anyone evaluate these?
    1. (integral sign) 2x+5(over) x(squared)+3x+6
    2. (integral sign) x(cubed)+5x(squared)+4(over) x(x(squared)+3x+2)
    ****over is equivalent to the division sign
  • Sep 25, 2010, 11:24 AM
    gashan
    these are the answers we got, can you confirm if they are correct. When I put x=0 in question 2, LHS and RHS worked out to be the same. But just want some confirmation.
    ∫ (2x + 5)/(x² + 3x + 6) dx
    = ∫ (2x + 3 + 2)/(x² + 3x + 6) dx
    = ∫ (2x + 3)/(x² + 3x + 6) dx + 2 ∫ dx/(x² + 3x + 6)
    = ∫ (2x + 3)/(x² + 3x + 6) dx + 2 ∫ dx/((x + 3/2)² + 15/4)
    = ∫ (2x + 3)/(x² + 3x + 6) dx + 2 ∫ dx/((x + 3/2)² + √(15)/2)
    = ln (x² + 3x + 6) + 2 (2/√15) arctan( 2(x + 3/2)/√15) + C
    = ln (x² + 3x + 6) + (4/√15) arctan( 2(x + 3/2)/√15) + C

    ∫ (x³ + 5x² + 4)/[x(x² + 3x + 2) ] dx
    = ∫ (x³ + 5x² + 4)/(x³ + 3x² + 2x) dx
    = ∫ (x³ + 3x² + 2x + 2x² - 2x + 4)/(x³ + 3x² + 2x) dx
    = ∫ (x³ + 3x² + 2x)/(x³ + 3x² + 2x) dx + ∫ (2x² - 2x + 4)/(x³ + 3x² + 2x) dx

    Let
    (2x² - 2x + 4)/(x³ + 3x² + 2x) = A/x + B/(x + 1) + C/(x + 2)
    Then
    2x² - 2x + 4 = A(x + 1)(x + 2) + Bx(x + 2) + Cx(x + 1)
    2x² - 2x + 4 = x²(A + B + C) + x(3A + 2B + C) + (2A)

    By comparing coefficients of different powers of x:
    2A = 4 ==> A = 2

    -2 = 3A + 2B + C ==> -2 = 6 + 2B + C ==> 2B + C = -8

    2 = A + B + C ==> 2 = 2 + B + C ==> B + C = 0 ==> B = -C

    ==> -2C + C = -8 ==> C = 8 ==> B = 8

    ==> (2x² - 2x + 4)/(x³ + 3x² + 2x) = 2/x + 8/(x + 1) - 8/(x + 2)

    continuation
    = ∫ (x³ + 3x² + 2x)/(x³ + 3x² + 2x) dx + ∫ (2x² - 2x + 4)/(x³ + 3x² + 2x) dx
    = ∫ 1 dx + ∫ (2/x + 8/(x + 1) - 8/(x + 2)) dx
    = x + 2ln|x| + 8ln|x + 1| - 8ln|x + 2| + C
  • Sep 25, 2010, 12:19 PM
    Unknown008

    Please, do not double post. Thank you.

    https://www.askmehelpdesk.com/math-s...ls-510669.html

    THREAD CLOSED
  • Sep 25, 2010, 12:59 PM
    gashan
    Comment on Unknown008's post
    I was trying to delete this, but I couldn't. I just started using this site recently, so I am still getting use to the options. Sorry about that

  • All times are GMT -7. The time now is 05:32 AM.