Results 1 to 15 of 15

Math Help - Integration Help

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    6

    Integration Help

    ∫▒x^3/(1+x^2 ) dx

    This is the question I have. It's quite a mind boggler (well, for me)

    It has to do with further substitution with integration.

    This is what I did, but I got the wrong answer in the back of the book.

    Let u = 1+x^2

    du/dx = 2x
    du/2x = dx

    ∫〖(x^3/u)(du/2x)〗

    ∫((u-1)/2u du)

    ∫〖1/2-1/2u〗

    =1/2 u- 1/2 ln⁡u

    =1/2(1+x^2 )-1/2 ln⁡(1+x^2 )+c

    I have a feeling it's something with my substitution
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by macho_maggot2001 View Post
    ∫▒x^3/(1+x^2 ) dx

    This is the question I have. It's quite a mind boggler (well, for me)

    It has to do with further substitution with integration.

    This is what I did, but I got the wrong answer in the back of the book.

    Let u = 1+x^2

    du/dx = 2x
    du/2x = dx

    ∫〖(x^3/u)(du/2x)〗

    ∫((u-1)/2u du)

    ∫〖1/2-1/2u〗

    =1/2 u- 1/2 ln⁡u

    =1/2(1+x^2 )-1/2 ln⁡(1+x^2 )+c

    I have a feeling it's something with my substitution
    Your substitution and work is fine - when your answer is differentiated the integrand is obtained, you've included the arbitrary constant of integration, so all is well.

    But remember that your answer includes an arbitrary constant (of integration) C ..... This is (probably) the clue that answers your question ......

    Note that \frac{x^3}{1 + x^2} = x - \frac{x}{1 + x^2}. When this is integrated, you get \frac{x^2}{2} - \frac{1}{2} \ln (1 + x^2) + K. This answer is also correct. Is it the book's answer? How can two apparently different answers be explained?? Well ......

    What happens if you replace the arbitrary constant C of your answer with the equally arbitrary K - \frac{1}{2} ....?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2007
    Posts
    6
    Ah....

    I get it now. It has to do with partial fractions. Interesting how the two are so different but have the same basic answer at the end

    Thank you so much!!
    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 macho_maggot2001 View Post
    Ah....

    I get it now. It has to do with partial fractions. Interesting how the two are so different but have the same basic answer at the end

    Thank you so much!!
    But do you understand that your answer and the book's answer are both correct - just found using different methods. Each answer differs by a constant, so they are both correct.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2007
    Posts
    6
    Oh, yes I see that.

    My next question has to again deal with u substitution, and I have no idea how to start it.

    It is:

    Integrate with respect to x:

    1/((x)(x^2+16)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by macho_maggot2001 View Post
    Oh, yes I see that.

    My next question has to again deal with u substitution, and I have no idea how to start it.

    It is:

    Integrate with respect to x:

    1/((x)(x^2+16)
    Use partial fractions to express \frac{1}{x(x^2 + 16)} in the form \frac{A}{x} + \frac{Bx + C}{x^2 + 16}, since x^2 + 16 is an irreducible quadratic. You will find that C = 0, which makes things pretty simple .....
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2007
    Posts
    6
    Why does it have to be Bx+C
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by macho_maggot2001 View Post
    Why does it have to be Bx+C
    Quote Originally Posted by mr fantastic View Post
    Use partial fractions to express \frac{1}{x(x^2 + 16)} in the form \frac{A}{x} + \frac{Bx + C}{x^2 + 16}, since x^2 + 16 is an irreducible quadratic. You will find that C = 0, which makes things pretty simple .....
    The stuff in red is why. Are you familiar with partial fraction decomposition?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    New questions in new topics.

    Quote Originally Posted by macho_maggot2001 View Post
    Integrate with respect to x:

    1/((x)(x^2+16)
    Substitute u=x^2+16, the rest follows easily.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Krizalid View Post
    New questions in new topics.


    Substitute u=x^2+16, the rest follows easily.
    The rest follows easily if the integrand is \frac{x^2 + 16}{x} ...... My reading of the question is that the integrand is \frac{1}{x(x^2 + 16)} .....
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    Quote Originally Posted by mr fantastic View Post
    My reading of the question is that the integrand is \frac{1}{x(x^2 + 16)} .....
    That was my reading of the question too, and the rest still follows easily.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Krizalid View Post
    That was my reading of the question too, and the rest still follows easily.
    For macho_maggot2001 , I'm not sure how getting and solving \frac{1}{2} \int \frac{1}{(u - 16) u} \, du is going to follow easily.

    By this I mean that macho_maggot2001 might understand 'follows easily' as implying that this substitution is somehow an easier road to take than the original proposed partial fraction decomposition. Arguably, it's not.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    We can avoid partial fraction descomposition, that's why I said "easily".

    \frac{1}<br />
{{u(u - 16)}} = \frac{1}<br />
{{16}} \cdot \frac{{u - (u - 16)}}<br />
{{u(u - 16)}} = \frac{1}<br />
{{16}}\left( {\frac{1}<br />
{{u - 16}} - \frac{1}<br />
{u}} \right).

    It useless to make the partial fraction descomposition since we can just tackle this with simple algebra.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Krizalid View Post
    We can avoid partial fraction descomposition, that's why I said "easily".

    \frac{1}<br />
{{u(u - 16)}} = \frac{1}<br />
{{16}} \cdot \frac{{u - (u - 16)}}<br />
{{u(u - 16)}} = \frac{1}<br />
{{16}}\left( {\frac{1}<br />
{{u - 16}} - \frac{1}<br />
{u}} \right).

    It useless to make the partial fraction descomposition since we can just tackle this with simple algebra.
    Don't get me wrong - I love it. And I loved your post here.

    But I'm of the opinion (and only macho_maggot2001 can confirm this) that the proposed substitution and what follows is not something that would follow easily for macho_maggot2001 (even after seeing it), or something that would serve him/her well for the next similar problem .....

    I just think, in context, the 'follows easily' is misleading to the member ...... My opinion, all comment welcome.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Sep 2007
    Posts
    6
    Oh, now I see. I believe the other post is a little easier on the eyes, but it all comes down to the same thing, right? It's all just partial fraction decomposition. I guess I didn't understand it at first because I have some inexperience with partial fractions. (I'm still in High School)

    Thank you guys so much!
    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