# Integration of $\sqrt{x+\sqrt{x^2+3x}}$ [closed]

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.

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 $${t}^{2}-{\frac {9}{4\,{t}^{2}+6}}+{\frac {27}{2\, \left( 2\,{t} ^{2}+3 \right) ^{2}}}$$

• That's the easy part, isn't it? Now you have to integrate that expression! (Or rather, $t \times$ that expression.) Nov 23, 2014 at 19:04
• the easy part? aha if you think so Jul 27, 2015 at 15:50
• the first is easy we obtain $\frac{t^3}{12}$ Jul 27, 2015 at 15:52
• the second one can be written in the form $$\frac{3}{8}\int\frac{dt}{\left(\sqrt\frac{2}{3}t\right)^2+1}$$ Jul 27, 2015 at 15:56
• Sep 3, 2017 at 19:46

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

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