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Three integrals

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data

    1)\ \int \frac{1}{(x^2+4)^3}\ \mbox{d}x

    2)\ \int \frac{x}{x^4+x^2+1}\ \mbox{d}x

    3)\ \int \frac{x e^x}{(x+1)^2}\ \mbox{d}x

    3. The attempt at a solution

    I got hints but I don't know how to use them.

    1) use trigonometric substitution and double angle formulas.
    Do I have to use [tex] u=cos( \theta) [/tex] and after substituting the double angle formulas what to do next?

    2) complete square in numerator
    But I'm missing a x^2 term!

    3) substitute y-1 and split integrand
    Split the term?
  2. jcsd
  3. Jun 11, 2008 #2
    1. I would say use x = tan(u)

    plug in values for x and dx.. and simplify. Integrate with respect to u and then convert from u to x
  4. Jun 11, 2008 #3
    That gives a [tex] \frac{1}{ \sec^4(u) }[/tex] does this makes life easier?
  5. Jun 11, 2008 #4
    I think the 2nd one can be solved as follows:

    [tex]{\int}xdx/(x^2+1) = 1/2\sdot{\int}dx^2/(x^2+1) [/tex] and the rest is a piece of cake or am I missing something?
  6. Jun 11, 2008 #5
    Yes a x^4 -term.
  7. Jun 11, 2008 #6


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    Hi dirk_mec1! :smile:

    1) Hint: [tex] \frac{1}{ \sec^4(u) }\,=\,cos^4u[/tex]

    2) Hint: No … you're missing a units term.

    3) Do the substitution anyway, and show us what you get. :smile:
  8. Jun 11, 2008 #7
    The last one:

    [tex] 3)\ \int \frac{x e^x}{(x+1)^2}\ \mbox{d}x = {\int}xde^x/(x+1)^2 = {\int} (x+1)de^x/(x+1)^2 - de^x/(x+1)^2 [/tex]

    The rest I leave for you. If I am not mistaken you should get [tex] e^x/(x+1) [/tex]

    Sorry for my LaTeX skills. I haven't figured it out yet
  9. Jun 11, 2008 #8
    Hi tiny tim,

    1) Oh now I see how to do the first one!
    2) what is a units term?
    3) working on it right now.
  10. Jun 11, 2008 #9


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    Sorry … I meant an ordinary number (with no x).

    Subtract a number from the denominator, to make a perfect square. :smile:

    (armis, nice LaTeX, but just give hints … unless the OP didn't follow the last hint! :smile: )
  11. Jun 11, 2008 #10
    yeah, sorry tiny-tim. Although often times it's kind of tricky to express my thoughts in english just verbaly but I'll try harder next time. Thanks god math notation is the same for everyone

    By the way I know it's not the right place to ask but how does one use the hide feature?
  12. Jun 11, 2008 #11
    You mean like: [tex] (x^2+1)^2 -x^2 [/tex]

    Use \frac{}{} for fractions
  13. Jun 11, 2008 #12
    oh, I see. Thanks
  14. Jun 11, 2008 #13


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    No … like [tex] (x^2+1)^2 - 3 [/tex] (but different, of course!) :smile:
    hmm … I was going to ask "what hide feature?" … and then I vaguely remembered seeing something about it … I think you type the full answer, but with a tag which hides it unless someone puts the mouse over it.

    But I've no idea how to operate it.

    You'd better ask someone more knowledgable than me! :redface:

    Anyway, it's good practice only giving hints! :wink:
  15. Jun 12, 2008 #14
    So you mean:

    [tex] \left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4} [/tex]

    So we get

    \int \frac{x}{x^4+x^2+1}\ \mbox{d}x = \int \frac{x}{ \left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4}}\ \mbox{d}x

    Now...what to do next?
    Last edited: Jun 12, 2008
  16. Jun 12, 2008 #15


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    :biggrin: Woohoo! :biggrin:
    Make the obvious substitution. :wink:
  17. Jun 12, 2008 #16

    Let [tex]u = x^2+\frac{1}{2} [/tex]


    \int \frac{x}{x^4+x^2+1}\ \mbox{d}x = \int \frac{x}{ \left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4}}\ \mbox{d}x = \frac{1}{2} \cdot \int \frac{1}{u^2+\frac{3}{4}}\ \mbox{d}x = \frac{1}{2} \arctan \left( \frac{u}{\frac{1}{2} \sqrt{3}} \right)\ +\ C

    Now put the u in terms of x back and we've got the answer. Is this okay Tim?
  18. Jun 12, 2008 #17


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    Yes, that's fine … except you've left out the (2/√3) outside the arctan.

    (See Hootenanny's Library entry, List of Standard Integrals)

    How did you get on with question 3? :smile:
  19. Jun 12, 2008 #18
    I am sorry to interrupt, but wasn't there a much easier way to solve the 2nd problem?

    [tex]{\int}\frac{xdx}{(x^2+1)} = 1/2\sdot{\int}\frac{dx^2}{(x^2+1)}[/tex]
  20. Jun 12, 2008 #19
    Whoops, you're right.

    I use the wiki-page :wink:

    \int \frac{x e^x}{(x+1)^2}\ \mbox{d}x = \frac{x e^x}{(x+1)^2} - \int e^x \frac{(x+1)^2-2x^2-2x}{(x+1)^4}

    This is getting nasty....

    By the way I didn't understand Armis' hint :biggrin:
    Last edited: Jun 12, 2008
  21. Jun 12, 2008 #20
    Where did you get this integral? This isn't the second integral I posted, right?
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