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Thread: Integration by Substitution

  1. #1
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    Integration by Substitution

    $\displaystyle \int \frac{3x-x^3}{x^4-6x^2 + 5} dx.$

    Let $\displaystyle u = x^4 -6x^2 + 5$
    $\displaystyle \frac{du}{dx} = 4x^3 -12x$
    $\displaystyle dx = \frac{du}{4x^3 - 12x}$

    then ?
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  2. #2
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    Re: Integration by Substitution

    Hello, mastermin346!

    $\displaystyle \displaystyle \int \frac{3x-x^3}{x^4-6x^2 + 5}\,dx$

    We have: .$\displaystyle \displaystyle -\int\frac{(x^3 - 3x)\,dx}{x^4-6x^2+5}$

    Let: $\displaystyle u \,=\,x^4-6x^2+5 \quad\Rightarrow\quad du \,=\,(4x^3 - 12x)\,dx \,=\,4(x^3-3x)dx$

    . . . . $\displaystyle (x^3-3x)dx \,=\,\tfrac{1}{4}du$


    Substitute: .$\displaystyle \displaystyle -\int\frac{\frac{1}{4}du}{u} \;=\;-\tfrac{1}{4}\int\frac{du}{u} \;=\; -\tfrac{1}{4}\ln|u|+C $


    Back-substitute: .$\displaystyle -\tfrac{1}{4}\ln\left|x^4-6x^2+5\right| + C$

    Thanks from mastermin346
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