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  1. MHB Master
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    #1
    $\tiny{206.07.05.88}$
    \begin{align*}
    \displaystyle
    I_{88}&=\int\frac{1}{(x+2)\sqrt{x^2+4x+3}} \, dx \\
    &=\int\frac{1}{(x+2)\sqrt{x^2+4x+3+1-1}} \, dx \\
    &=\int\frac{1}{(x+2)\sqrt{(x+2)^2-1}} \, dx \\
    u&=(x+2) \therefore du=dx \\
    I_{88}&=\int\frac{1}{u\sqrt{u^2+1}} du
    \end{align*}

    $\textit{so far ???}$

  2. MHB Journeyman
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    #2
    Quote Originally Posted by karush View Post
    $\tiny{206.07.05.88}$
    $\textit{so far ???}$

    So far so good.

    EDIT: See answer #7.
    Last edited by Fernando Revilla; January 11th, 2017 at 02:44.

  3. MHB Master
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    #3 Thread Author
    I tried next..

    $u=tan\theta \therefore du = sec^2 \theta d\theta$

    but it got difficult...

  4. MHB Journeyman
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    #4
    Quote Originally Posted by karush View Post
    I tried next..

    $u=tan\theta \therefore du = sec^2 \theta d\theta$

    but it got difficult...

    Better, $t=\dfrac{1}{u}$. Then, $$\displaystyle\int\frac{1}{u\sqrt{u^2+1}} du =\ldots =-\int\frac{1}{\sqrt{t^2+1}} dt=\ldots $$

    EDIT: See answer #7.
    Last edited by Fernando Revilla; January 11th, 2017 at 02:45.

  5. Perseverance
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    #5
    Quote Originally Posted by karush View Post
    $\tiny{206.07.05.88}$
    \begin{align*}
    \displaystyle
    I_{88}&=\int\frac{1}{(x+2)\sqrt{x^2+4x+3}} \, dx \\
    &=\int\frac{1}{(x+2)\sqrt{x^2+4x+3+1-1}} \, dx \\
    &=\int\frac{1}{(x+2)\sqrt{(x+2)^2-1}} \, dx \\
    u&=(x+2) \therefore du=dx \\
    I_{88}&=\int\frac{1}{u\sqrt{u^2+1}} du
    \end{align*}

    $\textit{so far ???}$
    Nope. You're ok up until

    $$\int\frac{1}{u\sqrt{u^2+1}}\text{ d}u$$

    That should be

    $$\int\frac{1}{u\sqrt{u^2-1}}\text{ d}u$$

    $$u=\cosh(z),\quad\text{ d}u=\sinh(z)\text{ d}z\quad z=\arcosh(u)$$

    $$\int\frac{\sinh(z)}{\cosh(z)\sqrt{\cosh^2(z)-1}}\text{ d}z=\int\frac{1}{\cosh(z)}\text{ d}z$$

    $$\int\frac{2}{e^z+e^{-z}}\text{ d}z=2\int\frac{e^z}{e^{2z}+1}\text{ d}z=2\arctan(e^z)+C$$

    $$z=\arcosh(u)=\log\left(u+\sqrt{u^2-1}\right)$$

    so, after back-subbing the rest of the way, we have

    $$\int\frac{1}{(x+2)\sqrt{x^2+4x+3}}\text{ d}x=2\arctan\left(x+2+\sqrt{(x+2)^2-1}\right)+C$$

  6. MHB Journeyman
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    #6
    Quote Originally Posted by greg1313 View Post
    Nope. You're ok up until

    $$\int\frac{1}{u\sqrt{u^2+1}}\text{ d}u$$
    That should be
    $$\int\frac{1}{u\sqrt{u^2-1}}\text{ d}u$$

    Sorry, I had a distraction error.

  7. MHB Master
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    #7
    Quote Originally Posted by karush View Post
    $\tiny{206.07.05.88}$
    \begin{align*}
    \displaystyle
    I_{88}&=\int\frac{1}{(x+2)\sqrt{x^2+4x+3}} \, dx \\
    &=\int\frac{1}{(x+2)\sqrt{x^2+4x+3+1-1}} \, dx \\
    &=\int\frac{1}{(x+2)\sqrt{(x+2)^2-1}} \, dx \\
    u&=(x+2) \therefore du=dx \\
    I_{88}&=\int\frac{1}{u\sqrt{u^2+1}} du
    \end{align*}

    $\textit{so far ???}$
    As has been pointed out, this should be

    $\displaystyle \begin{align*} \int{ \frac{1}{u\,\sqrt{u^2 - 1}}\,\mathrm{d}u} \end{align*}$

    From here

    $\displaystyle \begin{align*} \int{ \frac{1}{u\,\sqrt{u^2 - 1}}\,\mathrm{d}u} &= \frac{1}{2} \int{ \frac{2\,u}{u^2\,\sqrt{u^2 - 1}} } \end{align*}$

    Now let $\displaystyle \begin{align*} v = u^2 - 1 \implies \mathrm{d}v = 2\,u \end{align*}$ giving

    $\displaystyle \begin{align*} \frac{1}{2}\int{ \frac{1}{\left( v + 1\right) \,\sqrt{v}}\,\mathrm{d}v } &= \int{ \frac{1}{\left[ \left( \sqrt{v} \right) ^2 + 1 \right]\,2\,\sqrt{v}}\,\mathrm{d}v } \end{align*}$

    Now let $\displaystyle \begin{align*} w = \sqrt{v} \implies \mathrm{d}w = \frac{1}{2\,\sqrt{v}}\,\mathrm{d}v \end{align*}$ giving

    $\displaystyle \begin{align*} \int{ \frac{1}{\left[ \left( \sqrt{v} \right) ^2 + 1 \right] \,2\,\sqrt{v}}\,\mathrm{d}v } &= \int{ \frac{1}{ w^2 + 1}\,\mathrm{d}w } \\ &= \arctan{ \left( w \right) } + C \\ &= \arctan{ \left( \sqrt{v} \right) } + C \\ &= \arctan{ \left( \sqrt{ u^2 - 1 } \right) } + C \\ &= \arctan{ \left[ \sqrt{ \left( x + 2 \right) ^2 - 1 } \right] } + C \end{align*}$

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