Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Urgent Integration Help

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    18

    Urgent Integration Help

    I need to find the Integral of Sqrt(x^2+2x)dx. It is a Integration by Partial Fractions exercise.

    I don't even know where to start. Any help on where to start would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    I don't see the way of applying partial fractions here. Do we have a typo?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2008
    Posts
    18
    Quote Originally Posted by Krizalid View Post
    I don't see the way of applying partial fractions here. Do we have a typo?
    Not a typo. That's how it's on the book. Is there any other way to solve it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by djo201 View Post
    I need to find the Integral of Sqrt(x^2+2x)dx. It is a Integration by Partial Fractions exercise.

    I don't even know where to start. Any help on where to start would be greatly appreciated.
    Maybe a typo in the question ....? 1/(x^2 + 2x) .....?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Then integrate by parts or use a trig. substitution.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2008
    Posts
    18
    Quote Originally Posted by mr fantastic View Post
    Maybe a typo in the question ....? 1/(x^2 + 2x) .....?
    Not a typo there neither. It's the integral of the square root of (x^2 + 2x) dx

    Krizalid: If it isn't necessary I don't have to use trigonometric expressions. If it's the only way, please explain a bit further.

    EDIT: If this helps, the solution given by the book is:

    (1/2)[(x+1)Sqrt(x^2+2x)-ln|x+1+Sqrt(x^2+2x)|]+c
    Last edited by djo201; February 5th 2009 at 06:33 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2008
    Posts
    18
    Sorry for Double Post.

    I got confused. It's not from the Partial Fractions Section. It's a Trig. Substitution exercise.
    Last edited by djo201; February 5th 2009 at 06:58 PM.
    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
    14
    I know something was wrong.

    Well then, to perform a trig. substitution, we have x^2+2x=x^2+2x+1-1=(x+1)^2-1 so you need to put x+1=\sec\varphi.

    (I closed your duplicate thread on calculus section, there's no reason to make an identical one.)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Nov 2008
    Posts
    18
    Quote Originally Posted by Krizalid View Post
    I know something was wrong.

    Well then, to perform a trig. substitution, we have x^2+2x=x^2+2x+1-1=(x+1)^2-1 so you need to put x+1=\sec\varphi.

    (I closed your duplicate thread on calculus section, there's no reason to make an identical one.)
    Thanks, I made the trig. subs. and I'm at the point where I have
    Integral of sec^3(z)-sec(z). I'm not allowed to use the Table of Integrals. Any other way of moving on from that?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Yes, the first one is doable by parts and the second one requires some clever solution:

    \sec z=\frac{{{\sec }^{2}}z+\sec z\tan z}{\sec z+\tan z}.

    Note that by differentiating the denominator you get the numerator.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Nov 2008
    Posts
    18
    Quote Originally Posted by Krizalid View Post
    Yes, the first one is doable by parts and the second one requires some clever solution:

    \sec z=\frac{{{\sec }^{2}}z+\sec z\tan z}{\sec z+\tan z}.

    Note that by differentiating the denominator you get the numerator.

    Sorry for asking so many questions. But this problem is eating me. I'm trying to integrate by parts but can't find how. Do I use u=sec(z) or u=tan(z)?

    Again, sorry for asking and asking. And thanks for all the help.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    \int\sec^3z\,dz=\int\sec z\sec^2z\,dz=\int\sec z(\tan z)'\,dz.

    Strongly recommended lecture: make your partial integration faster. (See my signature.)

    ---

    Or if you want to use the classic way, put u=\sec z and dv=\sec^2z\,dz.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Nov 2008
    Posts
    18
    Quote Originally Posted by Krizalid View Post
    \int\sec^3z\,dz=\int\sec z\sec^2z\,dz=\int\sec z(\tan z)'\,dz.

    Strongly recommended lecture: make your partial integration faster. (See my signature.)

    ---

    Or if you want to use the classic way, put u=\sec z and dv=\sec^2z\,dz.
    I'm sure your going to kill me if you read this. I'm trying with partial integration but I'm stuck (AGAIN).

    I got to sec(z)tan(z)-Integral of sec(z)tan^2(z)

    Am I doing something wrong?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    No, I'm not gonna kill ya 'cause you're doing a good work. Now observe that,

    \sec z\tan^2z=\sec z(\sec^2z-1)=\sec^3z-\sec z.

    Is this familiar few posts ago?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Nov 2008
    Posts
    18
    Quote Originally Posted by Krizalid View Post
    No, I'm not gonna kill ya 'cause you're doing a good work. Now observe that,

    \sec z\tan^2z=\sec z(\sec^2z-1)=\sec^3z-\sec z.

    Is this familiar few posts ago?
    Yes, it's the same as before I started to integrate by parts. If I keep trying to integrate that it's going to keep repeating and repeating. What should I do?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Need urgent help! Integration in C0[0,1]
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 7th 2010, 04:16 PM
  2. integration - urgent help please
    Posted in the Calculus Forum
    Replies: 10
    Last Post: November 4th 2007, 09:15 AM
  3. integration - urgent help please
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 28th 2007, 05:24 AM
  4. integration - urgent help please
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 17th 2007, 07:53 PM
  5. integration urgent help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 14th 2006, 07:32 AM

Search Tags


/mathhelpforum @mathhelpforum