Beautiful Indefinite Integrals. 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.
 A: 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.
A: \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)
A: $$\int\dfrac{x^{4n}(1+x^{4n})}{1+x^2}dx$$
Why? Because from $0$ to $1$ they give good approximations of $\pi$. See this
A: $$\int (\sqrt {\tan x}+\sqrt{\cot x})dx=\sqrt 2\arctan\dfrac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt 2} +C$$
A: 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.
A: $$\int \left| \sin{ax} \right|\,dx = {2 \over a} \left\lfloor \frac{ax}{\pi} \right\rfloor - {1 \over a} \cos{\left( ax - \left\lfloor \frac{ax}{\pi} \right\rfloor \pi \right)} + C$$
$$\int \left|\cos {ax}\right|\,dx = {2 \over a} \left\lfloor \frac{ax}{\pi} + \frac12 \right\rfloor + {1 \over a} \sin{\left( ax - \left\lfloor \frac{ax}{\pi} + \frac12 \right\rfloor \pi \right)} + C$$
A: The Gaussian integral isn't indefinite but its derivation and answer are still remarkable:
$$\int_{-\infty}^{+\infty} e^{-x^2}\,dx = \sqrt{\pi}$$
