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I need to solve

$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$

I've chosen the substitution variables

$$u=\sqrt{x^2+2}$$ $$du=\frac{x}{\sqrt{x^2+2}}$$

However, I am completly stuck at

$$\int\sqrt{1+xu} dx$$

Which let me believe I've chosen wrong substitution variables.

I've then tried letting $u=x^2+2$ or simply $u=x$, but it does not help me at all solving it.

Would someone please give me an hint on this ?

Thanks.

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  • $\begingroup$ You've not substituted 'du' into the integral. $\endgroup$
    – Kugelblitz
    Mar 16 '15 at 1:38
  • $\begingroup$ I know, because I was not able to do it, which let me believe that the path I've taken was wrong. Could you confirm ? $\endgroup$
    – student
    Mar 16 '15 at 1:39
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    $\begingroup$ @MathLearner: (+1) for a problem that Maple and Wofram can't solve. $\endgroup$
    – Frieder
    Mar 16 '15 at 2:45
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    $\begingroup$ @MathLearner What book is this? $\endgroup$ Mar 16 '15 at 3:30
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    $\begingroup$ Substitute $x^2=2\sinh^2t$, and then, after simplifying the integrand, feed the new integral expression to the mathematical software of your choice. $\endgroup$
    – Lucian
    Mar 16 '15 at 4:40
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I thought about Wolfram and Maple too, needs a little help. But I was wrong.

$$\int {\sqrt {1 + x\sqrt {{x^2} + 2} } } dx = \int {\sqrt {1 + \frac{1}{2}\sqrt {{x^2} + 2} } } 2xdx$$

\begin{gathered} y = {x^2} + 2 \\ dy = 2xdx \\ \end{gathered}

\begin{gathered} \int {\sqrt {1 + \frac{1}{2}\sqrt y } } dy = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt y + 2} \right)^{3/2}}\left( {3\sqrt y - 4} \right) \\ = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt {{x^2} + 2} + 2} \right)^{3/2}}\left( {3\sqrt {{x^2} + 2} - 4} \right) \\ \end{gathered}

Now Wolfram: This is an identity:

$$\int {\sqrt {1 + \frac{1}{2}\sqrt y } } dy = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt y + 2} \right)^{3/2}}\left( {3\sqrt y - 4} \right)$$

Resubstituting gives me:

$$\frac{2}{{15}}\sqrt 2 {\left( {\sqrt y + 2} \right)^{3/2}}\left( {3\sqrt y - 4} \right) = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt {{x^2} + 2} + 2} \right)^{3/2}}\left( {3\sqrt {{x^2} + 2} - 4} \right)$$

Differentiation:

$$\frac{d}{{dx}}\left( {\frac{2}{{15}}\sqrt 2 {{\left( {\sqrt {{x^2} + 2} + 2} \right)}^{3/2}}\left( {3\sqrt {{x^2} + 2} - 4} \right)} \right) = \sqrt 2 x\sqrt {2 + \sqrt {{x^2} + 2} } $$`

And:

$$\sqrt 2 x\sqrt {2 + \sqrt {{x^2} + 2} } \ne \sqrt {1 + x\sqrt {{x^2} + 2} }$$

Very poor! For me. Made a mistake by myself.

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    $\begingroup$ Differentiating the solution presented the integral obtained is that of \begin{align} \int \sqrt{2 + \sqrt{x^{2} + 2}} \, dx \end{align} which is not the integral desired $\endgroup$
    – Leucippus
    Mar 16 '15 at 2:45
  • $\begingroup$ @Leucippus: You've got the wrong integrand! I missed an x and first 2 must be 1. $$\int {\sqrt {1 + x\sqrt {{x^2} + 2} } } dx$$ $\endgroup$
    – Frieder
    Mar 16 '15 at 2:51
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    $\begingroup$ $\int {\sqrt {1 + x\sqrt {{x^2} + 2} } } dx = \int {\sqrt {1 + \frac{1}{2}\sqrt {{x^2} + 2} } } 2xdx$ is wrong: you've taken the $x$ out of the square root incorrectly. $\endgroup$
    – Chappers
    Mar 16 '15 at 3:00
  • $\begingroup$ @Chappers: 100.000 thanks! What a shame. It's half past four in the morning (Germany). I should go to sleep. $\endgroup$
    – Frieder
    Mar 16 '15 at 3:24
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$x=\sqrt{2}\tan{u},dx=\sqrt{2}\sec^2{u}du,\sqrt{1+x\sqrt{x^2+2}}dx=\sqrt{2+4\tan{u}|\sec{u}|}\sec^2{u}du=\dfrac{\sqrt{2\cos^2{u} \pm 4\sin{u}}}{|\cos{u}|\cos^2{u}}du=\pm\dfrac{\sqrt{2\cos^2{u} \pm 4\sin{u}}}{cos^4{u}}d\sin{u}=\pm\dfrac{\sqrt{2(1-v^2) \pm 4v}}{(1-v^2)^2}dv ,v=\sin{u}$

consider case "+",$2(1-v^2)+4v=2(2-(v-1)^2)=2(2-y^2),y=v-1$ $\dfrac{\sqrt{2(1-v^2) +4v}}{(1-v^2)^2}dv=\dfrac{\sqrt{4-2y^2}}{(y(y+2))^2}dy$

$\dfrac{1}{y^2(y+2)^2}=\dfrac{1}{4(y+1)}\left(\dfrac{1}{y^2}-\dfrac{1}{(y+2)^2}\right)=\dfrac{1}{4y}\left(\dfrac{1}{y}-\dfrac{1}{(y+1)}\right)-\dfrac{1}{4(y+2)}\left(\dfrac{1}{y+1}-\dfrac{1}{(y+2)}\right)$

note :

$\dfrac{\sqrt{4-2y^2}}{y^2},\dfrac{\sqrt{4-2y^2}}{y}...$can be solved,so the problem can be solved.

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Substitute $x=\frac{2-t}{2\sqrt t}$. Then, $dx=-\frac {2+t}{4t\sqrt t}\>dt$ and

\begin{align} &\int\sqrt{1+x\sqrt{x^2+2}}\>dx \\ = & -\frac18 \int \frac{t+2}{t^2} \sqrt{ 4+4t-t^2}\>dt = -\frac18 \int \frac{\sqrt{ 4+4t-t^2}}{t-2}d\left( \frac{(t-2)^2}t\right)\\ =& -\frac{t-2}{8t }\sqrt{ 4+4t-t^2}- \int \frac1{t\sqrt{ 4+4t-t^2}}dt\\ =& -\frac{t-2}{8t }\sqrt{ 4+4t-t^2}+\frac12 \tanh^{-1} \frac{t+2}{\sqrt{ 4+4t-t^2}}+C \end{align}

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Consider the integral \begin{align} I = \int \sqrt{1 + x \sqrt{x^{2} + 2}} \, dx \end{align} Make the substitution $x = \sqrt{2} \, csch(t)$ to obtain the integral \begin{align} I = - \sqrt{2} \, \int \sqrt{1 + 2 \, csch(t) \, coth(t)} \cdot csch(t) \, coth(t) \, dt. \end{align} Now Wolfram can calculate the integral and provides the result \begin{align} - \frac{I}{\sqrt{2}} &= \frac{1}{2} \sinh(t) \sqrt{2 coth(t) csch(t)+1} \left[\frac{g(t)}{\sqrt{4 cosh(t)+cosh(2 t)-1}} - csch^2(t) \right] \end{align} where \begin{align} g(t) = \ln\left(\tanh^2\left(\frac{t}{2}\right)\right) - \ln\left(1 + \tanh^2\left(\frac{t}{2}\right) + \sqrt{-\tanh^4\left(\frac{t}{2}\right) + 2 \tanh^2\left( \frac{t}{2}\right) +1} \right) \end{align} By making a careful backward substitution the result becomes \begin{align} I &= \frac{1}{\sqrt{2}} \left[ x \sqrt{1+x\sqrt{x^{2}+2}} + \ln\left( 1 + \frac{(x + \sqrt{x^2 + 2})^{2}}{2} + \sqrt{\frac{(x + \sqrt{x^2 + 2})^{4}}{4} + (x + \sqrt{x^2 + 2})^{2} - 1} \, \right) \right] \end{align}

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