How can we calculate this integral:     $$I=\int\limits_0^1\frac{x^n-1}{\ln(x)}\,\mathrm dx$$   I believe that integral is equal to $\ln(n+1)$, but I don't lnow how to prove it.
124072
$$I(\alpha)=\int_0^1 \frac{x^\alpha-1}{\ln x}dx.$$  As the title says, if someone could solve this without using the differentiation under the integral sign technique, I would be very grateful.
31760
This integral is very closely related to the sophmores dream that states  $$ \int_0^1 x^{-x}\mathrm{d}x = \sum_{n=1}^\infty n^{-n} = 1.27\ldots $$  For example here   Now I want to bound the integral, and showing that is less that 2.  For the interval $[0,1]$ a good bound is rewriting it to $\exp(x\log x)$ and using the expansion   $$ 1 - x \log(x) + \frac12 (-x \log(x))^2$$  but how does one handle $[1,\infty)$ ? In this answer here     gives bounds to the integral, but they are not tight enough.. So to taste my question again, how does one prove that   $$ \int_0^\infty \frac{\mathrm{d}x}{x^x} \leq 2 $$
99383
Find the limit,  $$L=\lim_{n\to \infty}\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$  My try:  $$ \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}xdx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}(1-x)dx&lt; \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx&lt; \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}(1-x)dx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}xdx$$  Now taking the limit I get that,  $$\frac{1}{4}&lt;L&lt;\frac{3}{4}$$  But, how can I get the exact answer!!  This is  from the American Mathematical Monthly.
109728
Please help me to find a closed form for the following integral: $$\int_0^1\log\left(\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\right)\,{\mathrm d}x.$$  I was told it could be calculated in a closed form.
34545
I have tried to let$f(a)=\int_0^1\frac{\ln(1-ax)}{x} dx$, which means     $f'(x)=-a\int_0^1 \frac{1}{x(1-ax)} dx=a\int_0^1 \left[\frac{a}{ax-1}-\frac{1}{x}\right] dx$.   However, I am stuck at here.
92560
I am asked to integrate by parts $\int \ln(x) dx$. But I'm at a loss isn't there supposed to be two functions in the integral for you to be able to integrate by parts?
61825
Suppose $f$ is continuous on $[0,1]$ and   $$ \int\limits_0^1 f(x) x^n d x = 0 $$   for all $n=0,1,2,...$. Then, $f(x) = 0$  I was trying to use by parts  $$ 0 = f(x) \frac{x^{n+1}}{n+1} \bigg|_0^1 - \int\limits_0^1 \frac{x^{n+1}}{n+1} f(x) dx $$  Thus,  $$ \frac{f(1)}{n+1} = \int\limits_0^1 \frac{ x^{n+1} f(x) dx }{n+1} \implies f(1) = \int\limits_0^1 x^{n+1} f(x) dx $$  but here I am stuck. Maybe my idea is not the best one? How would you approach this problem?
57939
$$\lim_{n \to \infty }n\int_{0}^{1}\frac{x^{n}}{1+x+x^{n}}dx$$  I factorize $x^n$ then I tried with $u=1+x^{1-n}$ but I didn't get too far. Also I tried to get something from $0&lt;x&lt;1$. I know that if $x$ is from $(0,1)$ then $x^n$ tends to $0$ as $n \to \infty$.
95551
I'd like some help with the following integral:  $$\int_0^\infty \lfloor x \rfloor e^{-x}\mathrm dx .$$  Thanks.
926
How do I show that $\ln(x) = \lim_{n\to\infty} n (x^{1/n} - 1)$?  I ran into this identity on . I haven't been able to find any proof online and my efforts to get from $\ln(x) := \int_1^x \frac{\mathrm dt}t$ to that limit have been a failure.
92005
I recently got stuck on evaluating the following integral. I do not know an effective substitution to use. Could you please help me evaluate:  $$\int_{0}^{\infty} \ln\left(1 - e^{-x}\right) \,\mathrm dx $$
74165
Can anyone help to compute this limit, please? $\{nx\}$ denotes the fractional part of $nx$.  Tried to replace $nx$ by another variable and write an entire integral as a sum of integrals: $$\sum\limits_{i=0}^{n-1}\int\limits_{i}^{i+1} $$ but it seems not to work out.
118451
My question is, how can I evaluate the following integral?  $$\int\frac1{\sqrt[4]{1+x^4}}\mathrm dx$$  Thanks.
114177
We know that $$\int_0^1\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi}8\ln 2,$$ but how about $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx?$$Prove that $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}.$$
87553
How to figure out the following integral? I have not been able to solve it from some time. $$\int_0^1\frac{x^{2} -1}{\log x}\,dx$$
108021
Is there a way to calculate the improper integral $\int_0^\infty \big(\frac{\ln x}{x - 1}\big)^2 dx$?  What have I tried:  $$\int_0^\infty \Big(\frac{\ln x}{x - 1}\Big)^2 dx = \int_0^1 \Big(\frac{\ln x}{x - 1}\Big)^2 dx + \int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2 dx = \int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2dx + \int_1^\infty\Big(\frac{\ln t}{\big(\frac{1}{t} - 1\big)t}\Big)^2dt = 2\int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2dx$$  From the above statement we can conclude, that the integral does indeed converge, as $0 \leq (\frac{\ln x}{x - 1})^2 \leq \frac{C}{(x - 1)^{\frac{3}{2}}}$, for some constant $C$. However, this new form of the integral is not much helpful in finding its exact value.  I also tried substituting $z = \ln x$ , then $dz = \frac{dx}{x}$. Then the integral becomes reduced to $2\int_{0}^{\infty} \frac{z^2}{e^z(e^z - 1)^2}dz$, which, unfortunately, also does not seem to be any easier.
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.