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    #1
    $\textsf{evaluate}$
    \begin{align}
    \displaystyle
    {I}&={\int{\frac{x+2}{x^2+1}dx}}\\
    &=\int{\frac{x}{x^{2}{+1}}dx{\ +\ 2}\int{\frac{1}{x^{2}{+1}}}}{\ }dx\\
    u&=x^{2}+1 \therefore \frac{1}{2x}du=dx\\
    x&=\sqrt{u-1}\\
    \end{align}
    ...?

    $\textit{calculator answer.?}$

    $\dfrac{\ln\left(x^2+1\right)}{2}
    +2\arctan\left(x\right)+C$
    Last edited by karush; December 30th, 2016 at 00:56.

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    #2
    Since $du = 2x\, dx$, then $(1/2)du = dx$. So $x\, dx/(x^2 + 1) = (1/2)du$. Take it from there.

    As for the other integral, use a tangent substitution.

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    #3 Thread Author
    $\textsf{substitute $x=\tan(u)
    \therefore dx=\sec^2(u)$ }$
    \begin{align}
    \displaystyle
    &=2\int\frac{\tan\left({u}\right)}
    {\tan^2(u)+1}
    \cdot \sec^2(u )\, du
    =2\int \tan\left({u}\right) \, du
    =-2\ln\left({\cos\left({u}\right)}\right)+C
    \end{align}
    $\textsf{substitute u=arctan{x}}$

    $\textit{this doesn't seem to be headed towards the answer}$

    $\textit{calculator answer}$

    $\dfrac{\ln\left(x^2+1\right)}{2}
    +2\arctan\left(x\right)+C$

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    #4
    The tangent substitution was meant for the second integral

    $$2\int \frac{1}{x^2 + 1}\, dx$$

    and not the integral $\int x\, dx/(x^2 + 1)$. I already explained how to evaluate that integral with the $u$ you had already chose, namely, $u = x^2 + 1$.

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    #5 Thread Author
    so...
    $\textsf{solving}\\$
    \begin{align}
    \displaystyle
    I_a&=\int \frac{x}{x^{2}+1} \, dx \\
    u&=x^{2}+1 \therefore \frac{1}{2x}du=dx\\
    \textit{u substitution}\\
    &=\frac{1}{2} \int \frac{1}{u}du =\frac{\ln\left({u}\right)}{2}\\
    \textit{backsubstition }\\
    I_a&=\dfrac{\ln\left(x^2+1\right)}{2}
    \end{align}
    $\textsf{solving}\\$
    \begin{align}
    \displaystyle
    I_b &= 2 \int \frac{1}{x^{2}+1} \, dx\\
    x&=\tan(u) \therefore dx=\sec^2(u) du \\
    u&=\arctan(x) \\
    \textit{u substitution }\\
    I_b &= 2 \int \frac{1}{\tan^{2}u+1} \, \sec^2(u) du \\
    &=2 \int 1 du= 2u\\
    \textit{backsubstition}\\
    I_b&=2\arctan{(x)}
    \end{align}


    $\textit{$I_a + I_b +C = $}$


    $\dfrac{\ln\left(x^2+1\right)}{2}
    +2\arctan\left(x\right)+C$
    Last edited by karush; December 30th, 2016 at 16:15.

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    #6
    I would simply write:

    $ \displaystyle I=\int\frac{x+2}{x^2+1}\,dx=\frac{1}{2}\int \frac{2x}{x^2+1}\,dx+2\int \frac{1}{x^2+1}\,dx=\frac{1}{2}\ln(x^2+1)+2\arctan(x)+C$

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    #7
    Quote Originally Posted by karush View Post
    so...
    $\textsf{solving}\\$
    \begin{align}
    \displaystyle
    I_a&=\int \frac{x}{x^{2}+1} \, dx \\
    u&=x^{2}+1 \therefore \frac{1}{2x}du=dx\\
    \textit{u substitution}\\
    &=\frac{1}{2} \int \frac{1}{u}du =\frac{\ln\left({u}\right)}{2}\\
    \textit{backsubstition }\\
    I_a&=\dfrac{\ln\left(x^2+1\right)}{2}
    \end{align}
    $\textsf{solving}\\$
    \begin{align}
    \displaystyle
    I_b &= 2 \int \frac{1}{x^{2}+1} \, dx\\
    x&=\tan(u) \therefore dx=\sec^2(u) du \\
    u&=\arctan(x) \\
    \textit{u substitution }\\
    I_b &= 2 \int \frac{1}{\tan^{2}u+1} \, \sec^2(u) du \\
    &=2 \int 1 du= 2u\\
    \textit{backsubstition}\\
    I_b&=2\arctan{(x)}
    \end{align}


    $\textit{$I_a + I_b +C = $}$


    $\dfrac{\ln\left(x^2+1\right)}{2}
    +2\arctan\left(x\right)+C$


    Your work here is correct.

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