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Thread: Intergration: Substitution - Question

  1. #1
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    Intergration: Substitution - Question

    $\displaystyle
    \int \frac{x}{1+x^4} \ dx
    $

    I first started by setting $\displaystyle u = 1+x^4$ ,but then $\displaystyle du = 4x^3dx$ , doesn't work out with the $\displaystyle xdx$ 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 07:05 PM.
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  2. #2
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    Quote Originally Posted by tttcomrader
    $\displaystyle
    \int \frac{x}{1+x^2} \ dx
    $

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

    $\displaystyle
    \int \frac{x}{1+x^4} \ dx
    $

    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:

    $\displaystyle
    \int \frac{x}{1+x^4} \ dx
    $

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