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I faced the following indefinite integration problem: $$\int \sqrt{x+\sqrt{x^2+3x}}dx$$

This result by WolframAlpha suggests that there is an elementary way to compute this integration. But I don't know how to start. Any hints would be appreciated.

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we set $$t=\sqrt{x+\sqrt{x^2+3x}}$$ then by squaring, We get $$t^2-x=\sqrt{x^2+3x}$$ by squaring again, We get $$t^4-2t^2x=3x$$ thus $$x=\frac{t^4}{3+2t^2}$$ and $$dx=4\,{\frac {{t}^{3} \left( {t}^{2}+3 \right) }{ \left( 2\,{t}^{2}+3 \right) ^{2}}} dt$$ for the integration use that the integrand can written as $${t}^{2}-{\frac {9}{4\,{t}^{2}+6}}+{\frac {27}{2\, \left( 2\,{t} ^{2}+3 \right) ^{2}}} $$

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    $\begingroup$ That's the easy part, isn't it? Now you have to integrate that expression! (Or rather, $t \times$ that expression.) $\endgroup$ – TonyK Nov 23 '14 at 19:04
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    $\begingroup$ the easy part? aha if you think so $\endgroup$ – Dr. Sonnhard Graubner Jul 27 '15 at 15:50
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    $\begingroup$ the first is easy we obtain $\frac{t^3}{12}$ $\endgroup$ – Dr. Sonnhard Graubner Jul 27 '15 at 15:52
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    $\begingroup$ the second one can be written in the form $$\frac{3}{8}\int\frac{dt}{\left(\sqrt\frac{2}{3}t\right)^2+1}$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 27 '15 at 15:56
  • $\begingroup$ WolframAlpha can find partial fractions (link). $\endgroup$ – user236182 Sep 3 '17 at 19:46
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See the other answer. To find $\int \frac{27}{2(2t^2+3)^2}\, dt$, you can avoid using non-real numbers (notice that $2t^2+3=at^2+bt+c$, where $b^2-4ac<0$, has no real roots and no non-trivial factorization over the real numbers. We can't use partial fractions with real numbers).

See this link (link) (Wikipedia (link) also has some formulas), which shows the integration formulas over the real numbers of $\int \frac{1}{(x^2+bx+c)^n}\, dx$, $\int \frac{x}{(x^2+bx+c)^n}\, dx$, and see the examples there, in particular the example $\int \frac{1}{(x^2+1)^2}\, dx$. It generalizes. Notice that

$$\left(\frac{1}{\left(x^2+bx+c\right)^n}\right)'=\frac{-(2x+b)n}{\left(x^2+bx+c\right)^{n+1}}$$

$$\left(\frac{x}{\left(x^2+bx+c\right)^n}\right)'=\frac{x^2+bx+c-nx(2x+b)}{\left(x^2+bx+c\right)^{n+1}}$$

Use integration by parts, partial fractions. $$\int \frac{1}{2t^2+3}\, dt$$

$$\int u\, dv=uv-\int v\, du$$

$$u=\frac{1}{2t^2+3}$$

$$du=\frac{-4t}{(2t^2+3)^2}\, dt$$

$$dv=dt, v=t$$

$$\int \frac{1}{2t^2+3}\, dt=\frac{t}{2t^2+3}-$$

$$-\int \frac{-4t^2}{(2t^2+3)^2}\, dt$$

Use partial fractions. You could use WolframAlpha (link) if you want, but it's not needed.

$$\frac{-4t^2}{(2t^2+3)^2}=\frac{6}{(2t^2+3)^2}-\frac{2}{2t^2+3}$$

Now find $\int \frac{1}{(2t^2+3)^2}\, dt$ in terms of $$\int\frac{1}{2t^2+3}\, dt=\frac{1}{3}\sqrt{\frac{3}{2}}\int\frac{d\left(\sqrt{\frac{2}{3}}t\right)}{\left(\sqrt{\frac{2}{3}}t\right)^2+1}=$$

$$=\frac{1}{\sqrt{6}}\arctan\left(\sqrt{\frac{2}{3}}t\right)+C$$

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