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Math Help - Integration by Parts

  1. #1
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    Integration by Parts

    I need help finishing an integration by parts problem that I've already started. I think my substitutions for u and dv are good, but I'm still stuck. I've scanned a copy of my work and I attached it to this post. Can someone please take a look at it and tell me what I'm doing wrong and/or what I need to do next?

    Thank you.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zachb View Post
    I need help finishing an integration by parts problem that I've already started. I think my substitutions for u and dv are good, but I'm still stuck. I've scanned a copy of my work and I attached it to this post. Can someone please take a look at it and tell me what I'm doing wrong and/or what I need to do next?

    Thank you.
    \int \frac {2x}{2x + 1}~dx = \int \frac {2x + 1 - 1}{2x + 1}~dx

    = \int \frac {2x + 1}{2x + 1}~dx - \int \frac {1}{2x + 1}~dx

    = \int 1 ~dx - \int \frac {1}{2x + 1}~dx

    Can you take it from here?
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  3. #3
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    No, I don't understand what's going on in that first line you posted. Can you explain?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zachb View Post
    No, I don't understand what's going on in that first line you posted. Can you explain?
    it's just a common trick that is used. i added 1 and subtracted 1 in the numerator *that way, it's like adding zero, so i am not changing anything). doing that allows me to simplify the way i did. now i could split the fraction into two fractions. one simplifies to 1 which is easy to integrate, the other simplifies to a fraction that is easily taken care of by substitution
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  5. #5
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    Thanks. That's a neat trick. My professor didn't teach us that.
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  6. #6
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    Hey. I need more help. I used the same technique of adding 1 and subtracting 1, but this time it doesn't work out as nicely as it did for the last one.

    I posted an attachment of my work again. You should just ignore the last part because I don't think that it makes any sense.
    Attached Thumbnails Attached Thumbnails Integration by Parts-problem11-2-.jpg  
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  7. #7
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    Hello, zachb!

    \int \arctan(4t)\,dt

    You started correctly . . .
    . . and the integral is simpler than you think.

    We have: . \begin{array}{ccccccc}u & = & \arctan(4t) & \quad & dv & = & dt \\<br />
du & = & \frac{4}{1+16t^2}\,dt & \quad & v & = & t\end{array}

    Then we have: . t\!\cdot\!\arctan(4t) \:- \:4\int\frac{t}{1+16t^2}\,dt


    The new integral is of the form: . \int \frac{du}{u}

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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    apparently i gave you the impression that adding and subtracting 1 ALWAYS works. that is not the case, it depends on the particular integral. examine the integral i used that technique on and the integral you attempted to use it on, notice any differences? don't try to just mindlessly absorb and apply techniques, look for the reasoning behind them. as Soroban rightly pointed out, your last integral was an easy substitution problem. You do not have to always add and subtract 1 (in fact, in some cases you have to add and subtract 2, or 3, or 7 or whatever, or multiply and divide by the same thing etc). do what is called for. the above mentioned techniques are only applied when we want to change an ugly integral where none of our old techniques work, into a nice integral where at least one of our techniques work. the integral you attempted to use it on did not require it, it was already a "nice" integral, where one of our old techniques (substitution) works
    Last edited by Jhevon; September 10th 2007 at 03:36 AM.
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  9. #9
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    Quote Originally Posted by zachb View Post
    I need help finishing an integration by parts problem that I've already started. I think my substitutions for u and dv are good, but I'm still stuck. I've scanned a copy of my work and I attached it to this post. Can someone please take a look at it and tell me what I'm doing wrong and/or what I need to do next?

    Thank you.
    INT.[2x / (2x +1)]dx

    You could have just done the division so you could have not got lost with the tricks.

    2x divided by (2x +1)
    = 1, remainder -1
    = 1 -1/(2x+1)

    So,
    INT.[2x / (2x +1)]dx
    = INT.[1 -1/(2x+1)]dx
    = INT.[1]dx -INT.[1/(2x+1)]dx
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  10. #10
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    Quote Originally Posted by zachb View Post
    Hey. I need more help. I used the same technique of adding 1 and subtracting 1
    You're expected to know that this technique not always is applicable. It depends of the integral.

    Since you're gonna integrate \arctan4t, first set a little change of variables according to y=4t\implies dy=4\,dt, the integral becomes to

    \int\arctan4t\,dt=\frac14\int\arctan y\,dy, which it's easy to handle it through integration by parts.
    Last edited by Krizalid; December 6th 2007 at 03:33 PM.
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