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  1. #1
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    Intergrate

    intergrate (2x-3)/(x^2+16) dx

    Question is intergrate by partial fractions if so what am I not seeing complete the square in the denominator? Thanks for Checking out question people.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Bust2000 View Post
    intergrate (2x-3)/(x^2+16) dx

    Question is intergrate by partial fractions if so what am I not seeing complete the square in the denominator? Thanks for Checking out question people.
    ah, i don't think partial fractions is the way to go here.

    note that \int \frac {2x - 3}{x^2 + 16}~dx = \int \frac {2x}{x^2 + 16}~dx - \int \frac 3{x^2 + 16}~dx

    for the first, do a substitution u = x^2 + 16

    for the second, do not complete the square. instead, factor out the 16 from the denominator and then do a substitution t = \frac x4, and go for the arctangent integral
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  3. #3
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    Question

    Can you give me a little deminstration of what you are saying you are saying do a u substition of intergral uv -intergral v du. Sorry I need to work with my equation converter more it doesnt recognize intergrals still. Thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Bust2000 View Post
    Can you give me a little deminstration of what you are saying you are saying do a u substition of intergral uv -intergral v du.
    you are mixing this up with integration by parts, that is not what i am talking about. i am saying you must do integration by substitution. please look up this method, and try not to confuse the two

    Sorry I need to work with my equation converter more it doesnt recognize intergrals still. Thanks
    \int \frac 3{x^2 + 16}~dx = \frac 3{16} \int \frac 1{\frac {x^2}{16} + 1}~dx = \frac 3{16} \int \frac 1{\left( \frac x4 \right)^2 + 1} ~dx

    now, does that integral look familiar? it should. now do a substitution t = \frac x4 to see.
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  5. #5
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    Hello, Bust2000!

    Intergrate: . <br />
\int\frac{2x-3}{x^2+16}\, dx
    Make two integrals: . \int\frac{2x\,dx}{x^2+16} \;-\;3\int\frac{dx}{x^2+16}


    The first integral is:. \int\frac{2x\,dx}{x^2+16}

    . . Let u \,=\,x^2+16\quad\Rightarrow\quad du\,=\,2x\,dx

    . . Substitute: .[ \int\frac{du}{u} \;=\;\ln|u| + C\quad\Rightarrow\quad \ln(x^2+16) + C


    And the second integral is of the form: . \int\frac{du}{u^2+a^2}\;=\;\frac{1}{a}\arctan\left  (\frac{u}{a}\right) + C


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