Results 1 to 11 of 11

Math Help - Integration Help

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    25

    Integration Help

    I need to integrate 1/(16x^3-4x^2+4x-1)

    I first split the denominator into (4x-1)(4x^2+1) and used partial fraction decomposition to get:

    the integral of (4/5)/(4x-1)+((-4/5)x-(1/5))/(4x^2+1)

    I know how to solve the first segment of the integral, but I don't know how to integrate ((-4/5)x-(1/5))/(4x^2+1)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    311
    Thanks
    2
    Quote Originally Posted by Simon777 View Post
    I need to integrate 1/(16x^3-4x^2+4x-1)

    I first split the denominator into (4x-1)(4x^2+1) and used partial fraction decomposition to get:

    the integral of (4/5)/(4x-1)+((-4/5)x-(1/5))/(4x^2+1)

    I know how to solve the first segment of the integral, but I don't know how to integrate ((-4/5)x-(1/5))/(4x^2+1)
    Try splitting it into two rational expressions - the first with (-4/5)x in the numerator and the second with -(1/5) in the numerator.
    The first one will involve ln(4x^2+1),
    the second will need substitution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jul 2009
    Posts
    593
    Thanks
    4
    So you have:

    \int \frac{4}{5(4x-1)}-\frac{4x+1}{5(x^{2}+1)}dx \Rightarrow

    \int \frac{4}{5(4x-1)}-\frac{4x}{5(x^{2}+1)}+\frac{1}{5(x^{2}+1)}dx

     u = 4x-1; du = 4dx

    w = x^2+1; dw = 2xdx

    \int \frac{du}{5u}-\frac{2dw}{5w}+\frac{1}{5(x^{2}+1)} \Rightarrow

    \frac{ln|4x-1|-ln(x^2+1)^{2}+tan^{-1}(x)}{5}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    25
    Quote Originally Posted by ANDS! View Post
    So you have:

    \int \frac{4}{5(4x-1)}-\frac{4x+1}{5(x^{2}+1)}dx \Rightarrow
    With everything after the subtraction sign, why is it just (x^2+1)? Shouldn't it be (4x^2+1)? That was what the original problem was split into.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2010
    Posts
    25
    Quote Originally Posted by Debsta View Post
    Try splitting it into two rational expressions - the first with (-4/5)x in the numerator and the second with -(1/5) in the numerator.
    The first one will involve ln(4x^2+1),
    the second will need substitution.
    Thank you, I got a little farther with that advise, but what is the integral of:

    1/(4x^2+1)

    I can't do u substitution and I can't convert this into tan^-1 because of the 4.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Simon777 View Post
    Thank you, I got a little farther with that advise, but what is the integral of:

    1/(4x^2+1)

    I can't do u substitution and I can't convert this into tan^-1 because of the 4.
    edit: mistake. refer to skeeter's post
    Last edited by Defunkt; March 27th 2010 at 09:36 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,694
    Thanks
    450
    Quote Originally Posted by Simon777 View Post
    Thank you, I got a little farther with that advise, but what is the integral of:

    1/(4x^2+1)

    I can't do u substitution and I can't convert this into tan^-1 because of the 4.
    \frac{1}{4x^2+1} = \frac{1}{(2x)^2+1} = \frac{1}{2} \cdot \frac{2}{(2x)^2+1}

    \int \frac{1}{2} \cdot \frac{2}{(2x)^2+1} \, dx

    \frac{1}{2}\arctan(2x) + C
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Jul 2009
    Posts
    593
    Thanks
    4
    Quote Originally Posted by Simon777 View Post
    With everything after the subtraction sign, why is it just (x^2+1)? Shouldn't it be (4x^2+1)? That was what the original problem was split into.
    It is. Doesn't change using trig sub though. Even if it wasn't a perfect square.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2010
    Posts
    25
    Thank you all so much for the help. Can someone please check my answer?

    ln|((4x-1)^(1/5)) / ((4x^2+1)^(1/10))| - (1/10)arctan(2x) + C
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    311
    Thanks
    2
    Quote Originally Posted by Simon777 View Post
    Thank you all so much for the help. Can someone please check my answer?

    ln|((4x-1)^(1/5)) / ((4x^2+1)^(1/10))| - (1/10)arctan(2x) + C
    Yes that looks right. Just one thing, because of the yukky exponents in the first term, I would have written the first term as:
    (1/5) ln|(4x-1)/sqr(4x^2+1)|
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jan 2010
    Posts
    25
    Quote Originally Posted by Debsta View Post
    Yes that looks right. Just one thing, because of the yukky exponents in the first term, I would have written the first term as:
    (1/5) ln|(4x-1)/sqr(4x^2+1)|
    Thanks for taking the time to check that.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum