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Math Help - Integration help

  1. #1
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    Integration help

    I rewrote this as

    = (2x + 1)(x + 1)^4 is that right??? then i just combine them and integrate after??????????????????
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    Re: Integration help

    You need to rewrite them as \int (2x+1)\cdot (x+4)^{-\frac{1}{2}}\text{ dx} and use integration by parts, so choose u(x)=(x+4)^{-\frac{1}{2}} and v(x)=2x+1, then

    \int u(x)\cdot v(x) \text{ dx}=v(x)\cdot U(x)-\int v'(x)\cdot U(x)\text{ dx},

    where v'(x) is the differential of v(x) and U(x) is the antiderivative of u(x).

    Feel free to ask, if anything I just wrote need explanation
    Last edited by Zouzz; December 11th 2012 at 10:18 AM.
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    Re: Integration help

    yah it should be ^-1/2 but i made a mistake on putting ^4
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    Re: Integration help

    \int \frac{2x+1}{\sqrt{x+4}} \, dx

    u = x+4

    x = u - 4

    du = dx

    \int \frac{2(u-4) + 1}{\sqrt{u}} \, du

    \int 2u^{1/2} - 7u^{-1/2} \, du
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    Re: Integration help

    So i got this

    = (2)2(x + 4)^1/2

    is that right????
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    Re: Integration help

    Quote Originally Posted by asilvester635 View Post
    So i got this

    = (2)2(x + 4)^1/2

    is that right????
    check antiderivatives by taking the derivative ... do you get back to the integrand?
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    Re: Integration help

    oh i get it!!!!! yes integration by substitution..... i wasn't familiar with the word integration by parts that's why... THANK YOU SO MUCH
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    Re: Integration help

    Quote Originally Posted by asilvester635 View Post
    oh i get it!!!!! yes integration by substitution..... i wasn't familiar with the word integration by parts that's why... THANK YOU SO MUCH
    Integration by Parts is different to Substitution. Substitution is used as the opposite of the Chain Rule for derivatives, while Integration by Parts is used as the opposite of the Product Rule for derivatives.
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    Re: Integration help

    can anyone check if i got it right.

    i got...

    = 2(x + 4)^1/2 - 7(x + 4)^-1/2 + c
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  10. #10
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    Re: Integration help

    Quote Originally Posted by asilvester635 View Post
    can anyone check if i got it right.

    i got...

    = 2(x + 4)^1/2 - 7(x + 4)^-1/2 + c
    check it yourself ... take the derivative of your solution.
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  11. #11
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    Re: Integration help

    We know that \int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx

    More compactly:

    \int u \, dv=uv-\int v \, du.\!

    For \int \frac{(2x+1)}{\sqrt{x+4}}\;\;dx Let dv=\frac{1}{\sqrt{x+4}} and u=2x+1

    \therefore v = 2\sqrt{x+4}\;\; \text{ and } du = 2\;\;dx

    where v is the antiderivative of dv and du is the derivative of u




    \begin{align*}\int \frac{(2x+1)}{\sqrt{x+4}}\;\;dx  =& (2x+1)2\sqrt{x+4} - \int 4\sqrt{x+4} \;\;dx\\ =& 2(2x+1)\sqrt{x+4} -\frac{8}{3}(x+4)^{\frac{3}{2}}+C\\ =& \frac{2}{3}\sqrt{x+4}(2x-13)+C.......\text{[After simplification]}\end{align*}

    \therefore\int \frac{(2x+1)}{\sqrt{x+4}}\;\;dx  = \frac{2}{3}\sqrt{x+4}(2x-13)+ C

    Check:

    http://www.wolframalpha.com/input/?i=integration+(2*x%2B1)(x%2B4)^(-1%2F2)+dx
    Last edited by x3bnm; December 17th 2012 at 09:43 PM.
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