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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.

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closed as too broad by Antonio Vargas, Hans Engler, Daniel Rust, Grigory M, Arkamis May 15 at 19:42

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

    
Define "beautiful". –  Awesome May 15 at 17:50
1  
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 at 17:51
    
@mirgee actually, second one integrates pretty ugly too. –  Kaster May 15 at 17:53
    
I've always been very fond of the integrals of $\sec{x}$ and $\sec^3{x}$. –  David H May 15 at 17:53
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I believe this should be a community wiki though... –  Pound May 15 at 17:56

7 Answers 7

up vote 6 down vote accepted

\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)

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awesome work, thank you for posting this! –  userX May 16 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.

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

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

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

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

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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.

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I don't understand the reason of the downvote. '+1' –  user31782 Jun 3 at 16:20

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

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

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

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

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