# Beautiful Indefinite Integrals. [closed]

These are some of the integrals with beautiful solutions I came across-

$$\int \frac{x^2}{(x\sin x+\cos x)^2} dx$$

$$\int\frac {1}{\sin^3x+\cos^3x} dx$$

$$\int \frac{1}{x^4+1}dx$$

I'd love if you share some of the ones you came across.

## closed as too broad by Antonio Vargas, Hans Engler, Dan Rust, Grigory M, EmilyMay 15 '14 at 19:42

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• Define "beautiful". – evil999man May 15 '14 at 17:50
• I know perception of beauty is subjective, but what do you see as beautiful about the solution of the third integral, for example? – mirgee May 15 '14 at 17:51
• @mirgee actually, second one integrates pretty ugly too. – Kaster May 15 '14 at 17:53
• I've always been very fond of the integrals of $\sec{x}$ and $\sec^3{x}$. – David H May 15 '14 at 17:53
• I believe this should be a community wiki though... – Dair May 15 '14 at 17:56

\begin{align} I_1 & = \int \sqrt{ \sqrt{ x + 2\sqrt{2x-4} } + \sqrt{ x - 2\sqrt{2x-4} } } \,\mathrm{d}x \, , \quad x>4\\ I_2 & = \int \log( \log x) + \frac{2}{\log x} - \frac{1}{(\log x)^2} \mathrm{d}x \\ I_4 & = \int (1 + 2x^2) e^{x^2}\, \mathrm{d}x \\ I_5 & = \int \frac{\sqrt{x+\sqrt{x^2+1\,}\,}\,}{\sqrt{x^2+1\,}\,} \mathrm{d}x \\ I_6 & = \int \frac{2^x 3^x}{9^x - 4^x} \,\mathrm{d}x \end{align} \begin{align*} I_7 = \int \left( \frac{\arctan x}{x - \arctan x}\right)^2 \mathrm{d}x = \frac{1 + x \arctan x}{\arctan x - x} = \frac{1}{\tan (\beta - \tan \beta)}\,, \end{align*} where $x = \tan \tan \beta$ or $\beta = \arctan (\arctan x)$. $$I_6 = \int \frac{x^2+2x+1+ (3x+1)\sqrt{x+\ln x}}{x\,\sqrt{x+\ln x}(x+\sqrt{x+\ln x})}\mathrm{d}x = 2 (\sqrt{x+\ln x} + \ln(x+\sqrt{x+\ln x})) + C$$ I have a bunch more of these here, see p.68 for instance. (click on the problems for solution)

• awesome work, thank you for posting this! – userX May 16 '14 at 16:32

This isn't indefinite. But it's crazy

$$\int_0^{\pi/2} \frac{ d \theta}{\sqrt{a^2\cos^2\theta +b^2 \sin^2\theta }} = \frac{\pi}{2AGM(a,b)}$$

Where AGM is the arithmetic geometric mean.

$$\int\dfrac{x^{4n}(1+x^{4n})}{1+x^2}dx$$

Why? Because from $0$ to $1$ they give good approximations of $\pi$. See this

• cleverly constructed!! (+1) – Santosh Linkha May 15 '14 at 18:09

$$\int (\sqrt {\tan x}+\sqrt{\cot x})dx=\sqrt 2\arctan\dfrac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt 2} +C$$

Wolfram alpha gives time exceeded on this one :

$$\int\dfrac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx=\sqrt{t^2+2t-3}-\ln{(t+1+\sqrt{t^2+2t-3})}+\sqrt 3 \arcsin{\dfrac{t+5}{2(t+2)}}+C$$

where $t=x+\dfrac 1 x$

One does not simply integrate this.

• I don't understand the reason of the downvote. '+1' – user103816 Jun 3 '14 at 16:20

$$\int \left| \sin{ax} \right|\,dx = {2 \over a} \left\lfloor \frac{ax}{\pi} \right\rfloor - {1 \over a} \cos{\left( ax - \left\lfloor \frac{ax}{\pi} \right\rfloor \pi \right)} + C$$

$$\int \left|\cos {ax}\right|\,dx = {2 \over a} \left\lfloor \frac{ax}{\pi} + \frac12 \right\rfloor + {1 \over a} \sin{\left( ax - \left\lfloor \frac{ax}{\pi} + \frac12 \right\rfloor \pi \right)} + C$$

The Gaussian integral isn't indefinite but its derivation and answer are still remarkable:

$$\int_{-\infty}^{+\infty} e^{-x^2}\,dx = \sqrt{\pi}$$