How do we evaluate $\int\frac{1}{\ln{x}}dx$? Is it a special integral, or some simple function, or what substitution will help?
121772
What is the value of   $$\int\frac{1}{\log x}dx$$ I have tried many times, but failed everytime.   Can anyone help me out in solving this question.
48905
I would like to show that   $$ \int_{0}^{1} \frac{x-1}{\ln(x)} \mathrm dx=\ln2 $$  What annoys me is that $ x-1 $ is the numerator so the geometric power series is useless.  Any idea?
99407
How to do the following integral:  $$\int_{0}^1 \dfrac{x^a-1}{\log(x)}dx$$ where $a \geq 0$?  I was asked this question by a friend, and couldn't think of any substitution that works. Plugging in a=2,3, etc in Wolfram, I get values like $\log(a+1)$, which may be the right answer (for general $a$). Is there a simple way to calculate this integral?
58881
Evaluate the following integral $$\displaystyle I=\int_{0}^{1}\frac{x-1}{(x+1)(\ln x)} \mathrm{d}x $$ My work: I tried it by letting $\displaystyle I(a)=\int_{0}^{1}\frac{(x-1)x^a}{(x+1)(\ln x)} \mathrm{d}x$ and then $\displaystyle I'(a)=\int_{0}^{1}\frac{(x-1)x^a}{x+1} \mathrm{d}x$. Now $\displaystyle I'(a)=\int_{0}^{1}x^a \mathrm{d}x-\int_{0}^{1}\frac{2x^a}{x+1} \mathrm{d}x$ Now if $\displaystyle J(a)=\int_{0}^{1}\frac{x^a}{x+1}\mathrm{d}x$ , then by applying integration by parts, we get the reccurence relation $J(a)+J(a-1)=\dfrac{1}{a}$ and we can solve it then, but the thing is, we neet to find $I(0)$ ,so even if we compute $J(a)$, it wouldn't be defined at $0$ and so would $I(a)$, then how do I find $I'(a)$ by other method? I also tried the substitution $x \to \frac{1}{x}$, which yields $\displaystyle I=\int_{1}^{\infty}\frac{(x-1)}{(x+1)(\ln x)} \mathrm{d}x$ and when I saw their graphs, it clearly doesn't seem that the area under the graph of this function from $0$ to $1$ and from $1$ to $\infty$ are equal. I would appreciate if someone could  continue from my method and other solutions are also welcomed...
112982
Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{(2x)}  \, dx = \frac{\ln|2x|}{2}+ C$$   Check: By taking derivative we get $\frac{1}{x}$  Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{2x}  \, dx = \frac{1}{2}\int  \frac{1}{x}  \, dx =\frac{\ln|x|}{2}+ C$$  Check: By taking derivative we get $\frac{1}{x}$  Question: Is  $\int  \frac{1}{2x}  \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$?
114444
The inconsistency I see between mathematical subjects is really confusing me. I understand that it isn't possible for $e^x$ to be less than zero for real $x$, which is probably why they say that the integral is $\ln(|x|)$.  Before I ramble on too much, I just want to ask: Is there a set of guidelines to follow to help me choose whether to let the integral of $\frac{1}{x}$ equal to $\ln(x)$ or $\ln(|x|)$?  Thanks, Aralox
105432
Is there a way I can derive the value of the integral  $ \int_0^1 \ln(x)\ln(1-x)dx$  using the fact that  $\displaystyle\sum_{i=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$  ? (the actual value of the integral is $2-\frac{\pi^2}{6}$)  Thanks in advance
87933
$$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$  My Attempt:  $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt x\sqrt{1+x(1+x)}}$$  Replacing $x$ by $1-x$,we get  $$\int_{0}^{1}\frac{x}{2-x}.\frac{dx}{\sqrt{1-x}\sqrt{1+(1-x)(2-x)}}$$  Then I got stuck. Please help.
48353
Evaluate: $$ I(n) = \int_0^1\left(\frac{x-1}{x+1}\right)^n\frac{1}{\ln x} \,dx$$ For $n \in \mathbb N$.  I started by noticing that $\int_0^1 x^y \,dy = (x-1)/\ln x$ So I can convert this into a double integral as follows: $$I(n) = \int_0^1\int_0^1\frac{(x-1)^{n-1}x^y}{(x+1)^n}\,dy\,dx$$ Changing the order of integration and plugging in $x \rightarrow \cos 2\theta$, I get: $$I(n) = 2(-1)^n\int_0^1 \int_0^{\pi/4}\tan^{2n-1}\theta \cos^y 2\theta \,d\theta \,dy$$ I have a feeling that I can somehow transform the inner integral into a Beta function, but the $\pi/4$ and $2\theta$ are a bit irritating and hard to get rid of. Any methods to tackle this?
54941
$$ \int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2) $$  Anyone an idea on how to prove this?
42803
The original question was to evaluate:   $$\int_{0}^{1} \frac{1-x}{(1+x) \ln x}\,dx$$  Using the substitution $x=\tan^2 \theta $, I simplified it down to the integral $\int_{0}^{\frac{\pi}{4}} \frac{\tan \theta - \tan^3 \theta}{\ln \tan \theta}$.  From here, I am stuck and am not sure where to continue.
106102
How do I find the value of $$\int_{0}^{\infty} \frac{(\ln z)^2}{1+z^2}\mathrm{d}z$$ without using contour integration, - using the usual special functions, e.g., zeta/gamma/beta/etc.  Thank you,
43297
Can anyone give me a hint to evaluate this integral?  $$\int_0^\infty \frac{dx}{1+x^4}$$  I know it will involve the gamma function, but how?
69306
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 think that it is best solution to this problem.   I want to know if it can be solved by any other nice/elegant method.
68428
I have no idea how to find this integral:  $$\int_0^1\frac{(x-x^2)^4}{1+x^2}\ dx\ ?$$  Any help would be appreciated. Thanks in advance.
10157
I have stumbled onto an interesting integral$$\int_0^\infty \sin(x)\sin\left(\frac{1}{x}\right)dx$$ which I noticed graphically that it appears to be $1$, but I have no idea on how to evaluate it. Maybe it could be done with the use of Bessel Functions? Any help is appreciated.
88397
Evaluate integral $$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}$$ Well,I think we have $$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}=\frac{\pi}{e}$$  and  $$\int_{0}^{1}{x^{x}(1-x)^{1-x}\sin{\pi x}dx}=\frac{e\pi}{24}$$   With such nice result of these integral,why isn't worth to evaluate it?  I found a solution about the second one,but I wonder it will work for the first one   Note $$ S=\int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}-\int_{0}^{1}{(1-x)e^{(i\pi+\ln{x}-\ln{(1-x)})x}dx} $$ Let $t=\ln{x}-\ln{(1-x)}$,$x=\frac{e^{t}}{1+e^{t}}$ Thus \begin{align} S&amp;=\int_{-\infty}^{+\infty}{\frac{1}{e^{t}+1}e^{(i\pi+t)\frac{e^{t}}{1+e^t}}\frac{e^{t}}{(1+e^{t})^{2}}dt}\\   &amp;=\int_{-\infty+i\pi}^{-\infty-i\pi}{e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}}dt} \end{align} Due to $$ f(z)=e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}},\qquad D=\{Z\in C|-\pi\leq Im(z) \leq \pi\}$$ Therefore $res(f,0)=-\frac{e}{24}$when $z=0$ with $ \zeta_{R}=\gamma_{R}+o_{R}+\tau_{R}$ $$\oint_{\zeta_{R}}{f(z)dz}=-2\pi i\cdot res(f,0)=\frac{2i\pi e}{24}$$ because $$ \{z_{n}\}\subset D,\qquad |z_{n}|\rightarrow\infty $$ Therefore $$ 2S=2\lim_{R\rightarrow \infty}\int_{\gamma_{R}}{f(z)dz} $$ gives $$ \int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}=Im(S)=\frac{e\pi}{24} $$  My friend tian_275461 told me he use a simliar method to deal with the first one to obtain the result $\frac{\pi}{e}$,but I am not figure it out.