2021年1月20日 · I mean, when we take an integral and want it to be meaningful, we usually take definite integral, not indefinite integral. For $1/x$, the definite integral cannot be taken over an interval that contains 0, the two boundaries should be both positive or negative. So $\int_a^b \frac{1}{x}dx=\log(b/a)$, no mistake will be made. $\endgroup$ –

$$\int \frac{1}{1+x^4} \mathrm dx.$$ The integrand $\frac{1}{1+x^4}$ is a rational function (quotient of two polynomials), so I could solve the integral if I can find the partial fraction of $\frac{1}{1+x^4}$. But I failed to factorize $1+x^4$. Any other methods are also wellcome.

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$\begingroup$ These identities for $\int_0^1 x^{-x}\ dx$ and $\int_0^1 x^x\ dx$ are sometimes called the "sophomore's dream". Look that up on Wikipedia. $\endgroup$ – Robert Israel

2020年5月21日 · If you go to Flammable Maths's YouTube channel and scroll through some of his videos you see him solving the following integral: $$\int x^{dx}-1$$ he explains that this is a Product integral.

2021年10月3日 · The definition in many calculus textbooks is $$\ln(x) = \int_1^x \frac{1}{t} \, dt$$ I can imagine alternative definitions, but I would not want to guess which one you are assuming for this question. $\endgroup$

2021年10月1日 · A new number system called hyperreal number system was introduced by him, which contains infinitesimals and infinities. According to that, An infinitesimal 'dx' is a number which is smaller than all real numbers but greater than 0 i.e dx is infinitesimally closer to 0. It's denoted as st(dx)=0(where st denotes the standard part function).

Here's a bird's-eye view of the chain of arguments for how you get from the integral to the series: $$\begin{align} \int_{0}^{1}x^{-x}\mathrm{d}x&=\int_{0}^{1}\exp ...

2014年9月18日 · Evaluate the integral, $$ \int_{0}^{1} \ln(x)\ln(1-x)\,dx$$ I solved this problem, by writing power series and then calculating the series and found the answer to be $ 2 -\zeta(2) $, but I don't

2025年2月14日 · Does the following integral have a closed-form ?: \begin{equation} \int_{0}^{1}{\ln\left(\,x\,\right) \over 1 + x}\,\arccos\left(\,x\,\right) \,{\rm d}x \end{equation ...