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Math Help - indefinite integral

  1. #1
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    indefinite integral

    I'm totally lost on the following, need to find the indefinite integral...

    <br />
\int\frac{3x^{4}-5}{x^{2}}dx<br />

    <br />
\int\frac{4e^{x}}{5+e^{x}}dx<br />

    <br />
\int\frac{x+3}{x^{2}+6x+8}dx<br />
    Last edited by dasani; June 30th 2007 at 05:36 PM.
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  2. #2
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    Quote Originally Posted by dasani View Post
    I'm totally lost on the following,
    I am not surprised at that.
    What do you know about derivatives?
    Each of these requires that you know what was derived.
    Do you know the function was that gave these as derivatives?
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  3. #3
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    No there was no other info given by my teacher other than these these problems, and some of the basic rules used. I just can't figure out how to utilize those rules for these particular problems. The list of rules include the general power rule, etc...
    Last edited by dasani; June 30th 2007 at 05:38 PM.
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  4. #4
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    <br />
\int\frac{3x^{4}-5}{x^{2}}dx<br />
    Write this as,
    \int \frac{3x^4}{x^2} - \frac{5}{x^2} dx
    <br />
\int\frac{4e^{x}}{5+e^{x}}dx<br />
    Use the substitution t=5+e^x.
    <br />
\int\frac{x+3}{x^{2}+6x+8}dx<br />
    Use the substitution t=x^2+6x+8.
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  5. #5
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    Still confused on the substition method, and the first one seems like you re-wrote it...I'm still not sure what the next step is.
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  6. #6
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    Quote Originally Posted by dasani View Post
    <br />
\int\frac{3x^{4}-5}{x^{2}}dx<br />
    I thought you said you had been given the power rule?

    As TPH suggested:
    [tex]\int\frac{3x^{4}-5}{x^{2}}dx = \int \left ( 3x^2 - \frac{5}{x^2}\right ) dx <-- This is just simple division

    = \int (3x^2 - 5x^{-2}) dx

    Now for the power rule:
    = 3 \frac{1}{3}x^3 - 5 \frac{1}{-1}x^{-1} + C

    = x^3 + 5x^{-1} + C

    = x^3 + \frac{5}{x} + C

    -Dan
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dasani View Post
    <br />
\int\frac{4e^{x}}{5+e^{x}}dx<br />
    Again, as TPH suggested use t = 5 + e^x which implies
    dt = e^x dx

    dx = e^{-x}dt = \frac{1}{t - 5} dt

    So
    \int \frac{4e^{x}}{5+e^{x}}dx = \int \frac{4 \cdot (t - 5)}{t} \cdot \frac{1}{t - 5} dt = \int \frac{4}{t}~dt

    Now you take it from here.

    -Dan
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  8. #8
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    Quote Originally Posted by dasani View Post
    <br />
\int\frac{4e^{x}}{5+e^{x}}dx<br />
    \int\frac{4e^x}{5+e^x}~dx=4\int\frac{(5+e^x)'}{5+e  ^x}~dx=4\ln(5+e^x)+k

    Quote Originally Posted by dasani View Post
    <br />
\int\frac{x+3}{x^{2}+6x+8}dx<br />
     <br />
(x^2+6x+8)'=2(x+3)<br />
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  9. #9
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    You could always try the partial fraction thing for the third one.

    \frac{x+3}{x^{2}+6x+8}=\frac{1}{2(x+4)}+\frac{1}{2  (x+2)}

    \frac{1}{2}\int\frac{1}{x+4}dx+\frac{1}{2}\int\fra  c{1}{x+2}dx
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  10. #10
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    Thanks so much to all that replied, this was really helpful.
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