12k views

### Is there any integral for the Golden Ratio?

This is a curiosity. I was wondering about math important/famous constants, like $e$, $\pi$, $\gamma$ and obviously $\phi$. The first three ones are really well known, and there are lots of integrals ...
14k views

### Some users are mind bogglingly skilled at integration. How did they get there?

Looking through old problems, it is not difficult to see that some users are beyond incredible at computing integrals. It only took a couple seconds to dig up an example like this. Especially in a ...
7k views

### Integral $\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx$

Is there a closed form for the integral $$\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx.$$ I do not have a strong reason to be sure it exists, but I ...
7k views

### Evaluate $\int_0^{\pi/2}\frac{x^2\log^2{(\sin{x})}}{\sin^2x}dx$

How evaluate this integral? $$I=\int_0^{\pi/2}\frac{x^2\log^2{(\sin{x})}}{\sin^2x}\,dx$$ Note: $$\int_0^{\pi/2}\frac{x^2\log{(\sin x)}}{\sin^2x}dx=\pi\ln{2}-\frac{\pi}{2}\ln^22-\frac{\pi^3}{12}.$$
1k views

### Definite integrals with interesting results [closed]

I just stumbled across the fact that $\int_{-\infty}^{+\infty}{e^{-x^2}dx}=\sqrt{\pi}$. This intrigued my already-existing interest in integrals. It made me wonder, are there other integrals with ...
3k views

3k views

### Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$

I'm trying to evaluate this definite integral: $$\int_0^1\frac{\log(x) \log(1+x)}{\sqrt{1-x}} dx$$ It's clear that the result can be expressed in terms of derivatives of a hypergeometric function with ...
665 views

1k views

### What's the deal with integration?

So at uni we learned tricks and techniques for integration until cows came home. But to what end? Any/All definite integrals can be evaluated using numerical methods. Most integrals in application can ...
522 views

### Evaluate $\int_1^\infty \frac {dx}{x^3+1}$

I would like some help with the following integral. I would like to find a contour line to evaluate $$\int_1^\infty \frac {dx}{x^3+1}$$ So one can see that on any circumference it goes to $0$, but ...
765 views

### Integral $\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$

Calculate the following integral: $$\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$$ I am having trouble to calculate the integral. I tried ...
### How prove this $I=\int_{0}^{\infty}\frac{1}{x}\ln{\left(\frac{1+x}{1-x}\right)^2}dx=\pi^2$
Prove this $$I=\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\dfrac{1+x}{1-x}\right)^2}dx=\pi^2$$ My try: let $$I=\int_{0}^{\infty}\dfrac{2\ln{(1+x)}}{x}-\dfrac{2\ln{|(1-x)|}}{x}dx$$
### Does $\int_{-1}^1\frac{\arctan x}{\text{arctanh}\,x}\,\mathrm{d}x$ have a closed form?
$$\newcommand{\arctanh}{~\mathrm{arctanh}~}\newcommand{\sech}{~\mathrm{sech}~}$$ $$I=\int_{-1}^1\frac{\arctan x}{\arctanh x}\,\mathrm{d}x$$ Mathematica gives an approximate result of \$I=1....