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I have this integral:

$\int \sqrt{9 +3x^2} dx$

and I can't find way of solving it. Wolfram and Symbolab offer complicated solutions but for some reason I believe there is simpler one.

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  • $\begingroup$ You can use trigonometric substitution: en.wikipedia.org/wiki/Trigonometric_substitution $\endgroup$ – Joe Johnson 126 Dec 12 '16 at 18:12
  • $\begingroup$ Since it looks like $\int\sqrt{1+x^2}\,dx$, if an $\operatorname{arcsinh}$ (also known as $\ln\left(x+\sqrt{1+x^2}\right)$ ) appears I'm very much incline to believe them. $\endgroup$ – user228113 Dec 12 '16 at 18:12
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HINT: set $$x=\sqrt{3}\cosh(t)$$ and observe that $$\cosh(t)^2-\sinh(t)^2=1$$

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If you haven't learned how to use hyperbolic functions, the substitution $x = \sqrt 3 \tan t$ will work.

$dx = \sqrt 3 \sec^2 t$

$3 \sqrt 3 \int \sec^3 t \;dt$

Now we need to use integration by parts. I am going to set aside the $3 \sqrt 3$ constant for a little while.

$\int \sec^3 t \;dt\\ u = \sec t, dv = \sec^2 t\\ du = \sec t\tan t, v = \tan t$

$\sec t \tan t - \int \tan^2 t \sec t \;dt\\ \sec t \tan t - \int (\sec^2 t-1) \sec t \;dt$

And here is where you get tricky;

$\int \sec^3t\;dt = \sec t \tan t + \int \sec t\; dt - \int \sec^3 t \;dt\\ 2\int \sec^3t\;dt = \sec t \tan t + \int \sec t\; dt \\ \int \sec^3t\;dt = \frac 12 (\sec t \tan t + \ln |\sec t + \tan t|)$

Now we can bring back the constant, and reverse the substitution

$\frac {3\sqrt 3}{2} (x\sqrt{1+x^2} + \ln |\sqrt {1+x^2} + x|)$

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Hint...substitute $\sqrt{3}x=\sinh\theta$

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