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-   -   Evaluate the following integrals (https://www.askmehelpdesk.com/showthread.php?t=510669)

  • Sep 24, 2010, 10:59 PM
    gashan
    evaluate the following integrals
    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 24, 2010, 11:07 PM
    gashan
    can someone assist me with these integrals
    the s like sign e(raised to the power of 2x)+x(raised to the power of 2) cos3x dx. And the other question is
    the s sign 2x+1 divided by the square root of (x(raised to the power of 2)+3x+6)
  • Sep 25, 2010, 01:11 AM
    Unknown008

    1.

    This is close to the form:



    but you have f'(x) here = 2x + 3. So, break the fraction into two parts first.

    2. Try a long division first as the fraction isn't a simple fraction.

    3.

    Separate into two integrals and use by parts for the x^2cos(3x) one.

    4.

    Use integration by parts for this one.
  • Sep 25, 2010, 01:19 AM
    gashan
    Comment on Unknown008's post
    My best friend asked me to check on a solution for this question. I think he missed out a semester and is trying to teach himself from scratch. Would it be possible to give me the entire solution? Looking forward to your response
  • Sep 25, 2010, 04:31 AM
    galactus

    For the last one:



    Break it up into:



    For the first half, let


    See how? It whittles down to



    Now, continue with the second half.

    For that part, complete the square on the quadratic in the radical and then make an appropriate substitution.

    The first problem can be tackled by completing the square as well.
  • Sep 25, 2010, 05:43 AM
    gashan
    Comment on galactus's post
    For the second part, would it be equal to 1/4U+C?
  • Sep 25, 2010, 06:05 AM
    galactus

    No, it is a bit more involved than that.

    We start with:



    complete square:



    Let

    Then, we get:



    Now, try a trig sub perhaps to finish:



    Now, let

    Make the subs and it becomes:



    integrate and make the back substitutions. The solution involves ln
  • Sep 25, 2010, 06:28 AM
    gashan
    Comment on galactus's post
    OK cool. Thank you. Help greatly appreciated.
  • Sep 25, 2010, 11:27 AM
    gashan
    can you confirm if these answers that we got are correct. We placed x=0 in the second question, and both sides equate to the same. But just wanted 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, 12:03 PM
    Unknown008

    1.

    Complete the square;



    Sub u = x + 3/2.

    du = dx



    Use











    ->





    Okay, seems it agrees with your answer, good :)
  • Sep 25, 2010, 12:13 PM
    Unknown008

    2. B = -8, not 8

    Then, your answer becomes:

    x + 2ln|x| - 8ln|x + 1| - 8ln|x + 2| + C

    Good! :)

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