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  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post


    For the int. 2/x I got 2ln(x), for the int. x/x sq.+3 I got ln(x) and I'm not sure what to do for the int. 1/x sq. +3

    Can you give me a hint or show me what to do, please?

    Thank you
    your answer for \int \frac {x}{x^2 + 3}~dx is wrong. do it again. use substitution, u = x^2 + 3

    for \int \frac {1}{x^2 + 3}~dx we proceed as follows.

    note that: \frac {1}{x^2 + 3} = \frac {1}{3 \left( \frac {x^2}{3}  + 1 \right)}

    = \frac {1}{3} \cdot \frac {1}{ \left( \frac {x}{\sqrt {3}} \right)^2 + 1 }

    so, \int \frac {1}{x^2 + 3}~dx = \frac {1}{3} \int \frac {1}{ \left( \frac {x}{\sqrt {3}} \right)^2 + 1 }~dx

    now continue
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  2. #17
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    Thank you

    This is a u substitution too, right? Then you would do integration by parts, right?

    For the second interval I got 1/2ln|u|. Is that correct?

    Thank you
    Last edited by chocolatelover; September 3rd 2007 at 06:49 PM.
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  3. #18
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    Quote Originally Posted by Jhevon View Post
    \int \frac {1}{x^2 + 3}~dx = \frac {1}{3} \int \frac {1}{ \left( \frac {x}{\sqrt {3}} \right)^2 + 1 }~dx
    Wouldn't be faster if we take

    \int\frac1{x^2+3}\,dx=\int\frac1{x^2+\left(\sqrt3\  right)^2}=\frac1{\sqrt3}\arctan\frac x{\sqrt3}+k?

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  4. #19
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Wouldn't be faster if we take

    \int\frac1{x^2+3}\,dx=\int\frac1{x^2+\left(\sqrt3\  right)^2}=\frac1{\sqrt3}\arctan\frac x{\sqrt3}+k?

    yes, but that's basically memorizing a formula. my memory is horrible, so i always work things out the long way
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  5. #20
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    Quote Originally Posted by Jhevon View Post
    yes, but that's basically memorizing a formula.
    Not hard to memorize

    Actually, this helps when you're giving a test at the college, isn't?
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  6. #21
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    This is a u substitution too, right? Then you would do integration by parts, right?

    For the second interval I got 1/2ln|u|. Is that correct?
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  7. #22
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    This is a u substitution too, right? Then you would do integration by parts, right?

    For the second interval I got 1/2ln|u|. Is that correct?
    what integral are you talking about?

    there is no by parts here. regular substitution or a memorized formula will get you through

    Quote Originally Posted by Krizalid View Post
    Not hard to memorize
    that one is not hard to memorize, but it is one of many. if i memorize that one, i might as well memorize the one for arcsine, arccosine ...where does it end?!

    Actually, this helps when you're giving a test at the college, isn't?
    well, this level of algebraic manipulation doesn't slow me down really. at least not enough to warrant memorizing formulas that it is not the end of the world if i don't memorize them
    Last edited by ThePerfectHacker; September 3rd 2007 at 07:29 PM.
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  8. #23
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    For this one do I use integration of parts? Is the integral x/x sq. +3 1/2ln|u|?
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  9. #24
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post


    For this one do I use integration of parts? Is the integral x/x sq. +3 1/2ln|u|?
    it is not integration by parts. you use the substitution u = \frac {x}{\sqrt {3}} and you will get the answer that Krizalid got by using the formula. see the previous page in this thread

    the answer is \frac1{\sqrt3}\arctan\frac x{\sqrt3}+C
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  10. #25
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    Is du just dx then?

    So, it would 1/3int. udu=

    1/3int.u=

    1/3 u2/2

    that means that=

    1/3(x/sq. 3)2

    How does that equal 1/3arctanx/sq.3?
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  11. #26
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    Try to get it more generally

    \int\frac1{a^2+u^2}\,du

    Apply trig. substitution defined by u=a\tan\theta
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  12. #27
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    Can you show me what you mean? My book shows me how to do this type of problem, but it skips a lot of steps and I just started learning this.

    Thank you
    Last edited by chocolatelover; September 4th 2007 at 08:14 PM.
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  13. #28
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Can you show me what you mean? My book shows me how to do this type of problem, but it skips a lot of steps and I just started learning this.

    Thank you
    \int \frac{1}{a^2 + u^2} ~ du

    Let u = a \cdot tan(\theta) \implies du = a  \cdot sec^2(\theta) d \theta

    Thus
    \int \frac{1}{a^2 + u^2} ~ du = \int \frac{1}{a^2 + a^2tan^2(\theta)} \cdot (a \cdot sec^2(\theta)) d\theta

    = \int \frac{a}{a^2}\frac{sec^2(\theta)}{1 + tan^2(\theta)}~ d \theta

    Now, 1 + tan^2(\theta) = sec^2(\theta), which is the whole motivation for the form of the substitution, so
    = \frac{1}{a} \int \frac{sec^2(\theta)}{sec^2(\theta)}~d \theta

    = \frac{1}{a} \int ~d \theta

    = \frac{1}{a} \cdot \theta + C

    And u = a \cdot tan(\theta) \implies \theta = tan^{-1} \left ( \frac{u}{a} \right ), so finally:

    \int \frac{1}{a^2 + u^2} ~ du = \frac{1}{a} \cdot tan^{-1} \left ( \frac{u}{a} \right ) + C

    -Dan
    Last edited by topsquark; September 5th 2007 at 07:42 AM. Reason: This is what I get for not copying and pasting!
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  14. #29
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    Quote Originally Posted by topsquark View Post
    = \frac{1}{a} \cdot \theta + C

    And u = tan(\theta) \implies \theta = tan^{-1}(u), so finally:

    \int \frac{1}{a^2 + u^2} ~ du = \frac{1}{a} \cdot tan^{-1}(u) + C
    You missed something in the second line
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  15. #30
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    You missed something in the second line
    Two errors this morning and at least one last night! I'm off my game. Thanks for the spot. I fixed it in the original post.

    -Dan
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