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Thread: int x/(x^2+1)

  1. #1
    Boo
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    int x/(x^2+1)

    hello!

    I have to solve :
    $\displaystyle \int \frac{x}{x^2+1}dx=\frac{1}{2}ln|x^2+1|+C$

    well:$\displaystyle \int \frac{x}{x^2+1}dx=\frac{1}{2}\int\frac{2xdx}{x^2+1 }$

    if weknow that:
    $\displaystyle \int u^{-1}du=\int \frac{1}{u}du$ and $\displaystyle u=x^2+1$

    then what do we exactly do with 2xdx in the numerator? Dont we just have 2x too much???

    many thanks!
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  2. #2
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    Re: int x/(x^2+1)

    It's because when you substitute $\displaystyle \displaystyle \begin{align*} u = x^2 + 1 \end{align*}$, that means $\displaystyle \displaystyle \begin{align*} \frac{du}{dx} = 2x \end{align*}$. So

    $\displaystyle \displaystyle \begin{align*} \int{\frac{x}{x^2 + 1}\,dx} &= \frac{1}{2}\int{ \frac{1}{x^2 + 1}\cdot 2x \, dx} \\ &= \frac{1}{2} \int{ \frac{1}{u} \, \frac{du}{dx} \, dx } \\ &= \frac{1}{2}\int{ \frac{1}{u}\,du} \end{align*}$

    which you can now integrate.
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  3. #3
    Boo
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    Re: int x/(x^2+1)

    Thanks!
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