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Math Help - Working with Different Integration Techniques

  1. #1
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    Working with Different Integration Techniques

    Alright, so I've got a homework assignment that I've been working through and I've come across several problems that are tripping me up. If anyone could offer suggestions as to where I should start with these, it'd be greatly appreciated.

    #1) Evaluate the following integral:

    [xe^arctan(x) / (1+x^2)^(3/2)] dx

    #2) Evaluate the following integrals of rational functions:

    a) Integral of [1 / (4x^2+4x+1)] dx
    b) Integral of [1 / (4x^2+4x+5)] dx

    #3) I'm supposed to use the substitution t = tan(x/2) to convert the following integral into a rational function integral. (Put it in the lowest terms and then evaluate):

    Integral of:

    [1 / 1+sin(x)-cos(x)] dx

    Alright, so as far as what I've tried thus far:

    1) I didn't even know where to start...
    2) I figured that partial fractions would be the way to go here, so I broke up the integrals as per standard procedure and began to integrate, but when I finished that train of thought, I ended up getting something way different than the few people I asked (too long to even bother posting here)...
    3) Here, I was supposed to use a similar substitution as was used in a previous problem, but when I did, I ended up getting another weird answer, so I figured maybe some fresh eyes without any background info might be helpful to see something that I was missing...

    Thanks again.

    -B

    P.S. - I'm not digging for free help here, I just wanted some pointers on what I can do to start these and actually end up with the correct answer...
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  2. #2
    Member Mentia's Avatar
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    Bellingham, WA
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    I can help with the first one:

    Use the substitution: x = \tan u or u = \arctan x

    Then,

    du = \frac{1}{1+x^2}dx but also dx = \sec ^2 (u)du

    Apparently,

    (1+x^2) = \sec ^2 u

    Then we have:

    \frac{e^{\tan ^{-1}(x)} x}{\left(x^2+1\right)^{3/2}}dx = \frac{\tan (u) e^{u}}{\sec u}du = \sin (u) e^u du

    If you know the integral of \sin (u) e^u du then you are good to go. Otherwise you can split up sine as:

    \sin u = \frac{e^{iu}-e^{-iu}}{2i} and proceed from there.
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  3. #3
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    Alright, I think I see where you're going with this. Thanks for the help...

    Any ideas as to where I should begin with the others?

    -B
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  4. #4
    Member Mentia's Avatar
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    For #2, notice that the denominator in a) is just (1+2x)^2, so this one turns out to be really simple.
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