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Math Help - Intergration: Substitution - Question

  1. #1
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    Intergration: Substitution - Question

    <br />
\int \frac{x}{1+x^4} \ dx<br />

    I first started by setting u = 1+x^4 ,but then du = 4x^3dx , doesn't work out with the xdx that is left.

    Any ideas?

    By the way, I had finished my Spring Semester, with the help of this forum, I was able to pull As in my Calculus I and Linear Algebra. Just wanna thank you all!!!
    Last edited by tttcomrader; May 25th 2006 at 08:05 PM.
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  2. #2
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    Quote Originally Posted by tttcomrader
    <br />
\int \frac{x}{1+x^2} \ dx<br />

    I first started by setting u = 1+x^4 ,but then du = 4x^3dx , doesn't work out with the xdx that is left.
    Let,
    u=1+x^2 then, \frac{du}{dx}=2x
    Therefore, express integral as,
    \frac{1}{2}\int \frac{1}{1+x^2}\cdot (2x) dx
    Substitute,
    \frac{1}{2}\int \frac{1}{u}\frac{du}{dx} dx=\frac{1}{2}\int \frac{1}{u} du=\frac{1}{2}\ln|u|+CThus,
    \frac{1}{2}\ln |1+x^2|+C
    but, 1+x^2>0 therefore,
    \frac{1}{2}\ln (1+x^2)+C
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  3. #3
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    Oh, sorry, I made a mistake, the problem should be:

    <br />
\int \frac{x}{1+x^4} \ dx<br />

    Is to the 4th power instead of 2nd.
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  4. #4
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    Quote Originally Posted by tttcomrader
    Oh, sorry, I made a mistake, the problem should be:

    <br />
\int \frac{x}{1+x^4} \ dx<br />

    Is to the 4th power instead of 2nd.
    You have,
    \int \frac{x}{1+x^4}dx
    Let,
    u=x^2 then, \frac{du}{dx}=2x
    Express integrand as,
    \frac{1}{2}\int \frac{1}{1+(x^2)^2}\cdot (2x) dx and substitute,
    \frac{1}{2}\int \frac{1}{1+u^2}\frac{du}{dx}dx=\frac{1}{2}\int \frac{du}{1+u^2}=\frac{1}{2}\tan^{-1}(u)+C
    Substitute back,
    \frac{1}{2}\tan^{-1}(x^2)+C
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