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Math Help - how to solve this root integral

  1. #1
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    how to solve this root integral

    i tried this:
    <br />
\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}=\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}*\frac{1+\sqrt{x+1}}{1+\sqrt{x+1}}*\frac{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}<br />
    but when i got read of 2 roots i got another two roots which are more complicated
    ??
    Last edited by transgalactic; February 18th 2009 at 09:07 AM.
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    i tried this:
    <br />
\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}=\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}*\frac{1+\sqrt{x+1}}{1+\sqrt{x+1}}*\frac{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}<br />
    but when i got read of 2 roots i got another two roots which are more complicated
    ??
    Try x = u^6-1
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  3. #3
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    Hello, transgalactic!

    \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}\,dx
    Danny has the best idea . . .


    Let: u^6 \:=\:x+1 \quad\Rightarrow\quad x \:=\:u^6 - 1\quad\Rightarrow\quad dx \:=\:6u^5\,du

    . . Then: . \sqrt{x+1} \:=\:u^3,\;\sqrt[3]{x+1} \:=\:u^2


    Substitute: . \int\frac{1-u^3}{1+u^2}(6u^5\,du)  \;=\;6\int\frac{u^5 - u^8}{1+u^2}\,du


    Long division: . 6\int\bigg[-u^6 + u^4 + u^3 - u^2 - u + 1 + \frac{u-1}{u^2+1}\bigg]\,du


    Can you finish it now?

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  4. #4
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    <br />
\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}dx=\int \frac{(1-u^3)6u^5}{1+u^2}du=\int \frac{6u^5-6u^8}{1+u^2}du=<br />
    <br />
\int -6u^6+6u^3-6u+\frac{6u}{1+u^2}du=..<br />
    did i devided the polynomials correctly??
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  5. #5
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    Quote Originally Posted by transgalactic View Post
    <br />
\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}dx=\int \frac{(1-u^3)6u^5}{1+u^2}du=\int \frac{6u^5-6u^8}{1+u^2}du=<br />
    <br />
\int -6u^6+6u^3-6u+\frac{6u}{1+u^2}du=..<br />
    did i devided the polynomials correctly??
    as Soroban did
    first factor out -6 then you'll have:
    u^2+1)=u^6-u^4-u^3+u^2+u-1-\frac{u-1}{u^2+1}" alt="(u^8-u^5)u^2+1)=u^6-u^4-u^3+u^2+u-1-\frac{u-1}{u^2+1}" />
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  6. #6
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    i got a different remainder:
    -6u^6:
    <br />
 \frac{-6u^8+6u^5}{u^2+1} \\<br />
    <br />
-6u^8+6u^5-(-6u^8-6u^6)=6u^6+6u^5\\<br />
    6u^4:
    <br /> <br />
6u^6+6u^5-(6u^6+6u^4)=6u^5-6u^4<br />
    6u^3:
    <br /> <br />
6u^5-6u^4-(6u^5+6u^3)=-6u^4-6u^3\\<br />
    -6u^2:
    <br /> <br />
-6u^4-6u^3-(-6u^4-6u^2)=-6u^3+6u^2\\<br />
    -6u:
    <br /> <br />
-6u^3+6u^2-(-6u^3-6u)=6u^2+6u\\<br />
    6\\
    <br />
6u^2+6u-(6u^2+6)=6+6u<br />
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  7. #7
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    \int{\frac{u^{5}-u^{8}}{1+u^{2}}\,du}=\int{\frac{u^{5}}{1+u^{2}}\,d  u}-\int{\frac{u^{8}}{1+u^{2}}\,du}.

    Put v=1+u^2 for the first integral, and

    \frac{u^{8}}{1+u^{2}}=\frac{u^{8}-1+1}{1+u^{2}}=\frac{\left( u^{4}+1 \right)\left( u^{2}+1 \right)\left( u^{2}-1 \right)+1}{1+u^{2}}=\left( u^{4}+1 \right)\left( u^{2}-1 \right)\,+\,\frac{1}{1+u^{2}},

    which is very easy to integrate.
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