Results 1 to 9 of 9

Math Help - natural logarithm integrals

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    Thumbs down natural logarithm integrals

    Help im stuck with this
    \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx


     \int \frac{3x^5}{x^3+1} - \frac{2x^3}{x^3+1}+ \frac{5x^2}{x^3+1}-\frac{2}{x^3+1} dx
    ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Help im stuck with this
    \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx


     \int \frac{3x^5}{x^3+1} - \frac{2x^3}{x^3+1}+ \frac{5x^2}{x^3+1}-\frac{2}{x^3+1} dx
    ?
    Hello,

    do long division first:
    Code:
                                                    2x^2
     (3x^3 - 2x^3 + 5x^2 - 2)  (x^3+1= 3x^2 - 2 + -------
    -(3x^5        + 3x^2)                           x^3+1
    ----------------------
            -2x^3 + 2x^2
          -(-2x^3         -2)
           ------------------
                    2x^2
    Now use integration by substitution. Let u = x^3+1 \Longrightarrow \frac{du}{dx} = 3x^2

    \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{3x^2}{x^3+1} \right)dx
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381
    but quickmath has different answer
    QuickMath Automatic Math Solutions
    anyway the only problem i have here is the long division...
    how did it get through quickmath answer?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    but quickmath has different answer
    QuickMath Automatic Math Solutions
    anyway the only problem i have here is the long division...
    how did it get through quickmath answer?
    Hello,

    the denominator you used with quickmath is not the same you published with your problem
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    but quickmath has different answer
    QuickMath Automatic Math Solutions
    anyway the only problem i have here is the long division...
    how did it get through quickmath answer?
    Hello,

    I've tried to do long division with quickmath but didn't get a reasonable result. (I assume that I don't know the appropriate command).

    If you are allowed to use a TI89 then you can do long divisions by propFrac(...). Instead of the three dots write the fraction. You find this command in the F2 Algebra menu.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    but quickmath has different answer
    QuickMath Automatic Math Solutions
    anyway the only problem i have here is the long division...
    how did it get through quickmath answer?
    Quote Originally Posted by earboth View Post
    Hello,

    do long division first:
    \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{3x^2}{x^3+1} \right)dx
    It isn't the same! Earboth's solution didn't take into account that x^3 + 1 = (x + 1)(x^2 - x + 1). As it happens:
    \frac{3x^2}{x^3 + 1} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1}

    = \frac{Ax^2 - Ax + A + Bx^2 + (B + C)x + C}{x^3 + 1}

    = \frac{(A + B)x^2 + (-A + B + C)x + (A + C)}{x^3} = \frac{3x^2}{x^3 + 1}

    Thus
    A + B = 3
    -A + B + C = 0
    A + C = 0

    Gives:
    A = 1
    B = 2
    C = -1

    So
    \frac{3x^2}{x^3 + 1} = \frac{1}{x + 1} + \frac{2x - 1}{x^2 - x + 1}

    Which is not the same as QuickMath's answer. But the final partial fraction split should be done. So
    \frac{3x^5 - 2x^3 + 5x^2 - 2}{x^3 + 1} = 3x^2 - 2 + \frac{1}{x + 1} + \frac{2x - 1}{x^2 - x + 1}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by earboth View Post
    Hello,

    do long division first:
    Code:
                                                    2x^2
     (3x^3 - 2x^3 + 5x^2 - 2)  (x^3+1= 3x^2 - 2 + -------
    -(3x^5        + 3x^2)                           x^3+1
    ----------------------
            -2x^3 + 2x^2
          -(-2x^3         -2)
           ------------------
                    2x^2
    Now use integration by substitution. Let u = x^3+1 \Longrightarrow \frac{du}{dx} = 3x^2

    \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{3x^2}{x^3+1} \right)dx
    You have a transcription error at one point you have:

    \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{2x^2}{x^3+1} \right)dx

    then later you have:

    \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{3x^2}{x^3+1} \right)dx

    The first is correct, but the latter is not.

    Earboth's original decomposition is correct, and there is no need to use
    partial fractions after this as we now have the sum of a polynomial and
    a rational function who's numerator is a numeric multiple of the derivative
    of the denominator, and so its integral can be written down immediately.

    RonL
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Help im stuck with this

    \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx
    \begin{aligned}\int\frac{3x^5-2x^3+5x^2-2}{x^3+1}~dx&=3\int\frac{x^2(x^3+1)}{x^3+1}~dx-2\int{dx}+\int\frac{2x^2}{x^3+1}~dx\\&=3\int{x^2}~  dx-2x+\frac23\int\frac{(x^3+1)'}{x^3+1}~dx\\&=x^3-2x+\frac23\ln\left|x^3+1\right|+k,~~k\in\mathbb{R}  ~\blacksquare\end{aligned}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    Thumbs up

    thank you all
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Primitives of the natural logarithm
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 28th 2009, 10:26 PM
  2. Derivative with natural logarithm
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 5th 2009, 03:01 PM
  3. natural logarithm
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: June 15th 2009, 07:25 PM
  4. Natural logarithm
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 7th 2009, 07:38 PM
  5. Natural Logarithm
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 4th 2008, 01:08 AM

Search Tags


/mathhelpforum @mathhelpforum