How to compute the integral $\int^{\pi/2}_0\ln(1+\tan\theta)d\theta$. If we let $t=\tan\theta$, then the integral becomes to  $$\int^{\pi/2}_0\ln(1+\tan\theta)d\theta=\int_0^\infty\frac{\ln(1+t)}{1+t^2}dt$$.  Can we calculate this integral explicitly?
115282
Can this integral be solved with contour integral or by some application of residue theorem? $$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$  It has two poles at $\pm i$ and  branch point of $-1$ while the integral is to be evaluated from $0\to \infty$. How to get $\text{Catalan Constant}$? Please give some hints.
33350
Some time ago I came across  to the following integral:    $$I=\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$$ What are the hints on how to compute this integral?
28612
I know $\int \tan^{1/3}\theta  d \theta$ is a non integrable but wolfram alpha does it easily. And can anyone explain how a fuction is non integrable? My argument is that there has to be a function which represents the area under a graph. It's just that we do not know it. Please shed a bit light on it. Also, how can we solve $\int \tan^{1/3}\theta  d \theta$?
75155
So I came across this question on brilliant.org and didn't know how to go about it: $$\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $$  I tried to complete the squares in the denominator and then use a trigonometric substitution (with tan) but I didn't get anything beyond that.   How do I do it? Do we need to differentiate under the integral sign here?
32720
$${\dfrac{1}{\pi^2}}\int_{0}^{\infty}\dfrac{{(\ln{x}})^2}{\sqrt{x}{(1-x)^2}} \mathrm d x$$   I tried substituting $1/x$ for $x$ but the the only change in the integral is that the $\sqrt{x}$ moves in the numerator from the denominator. I don't understand what to substitute. $\tan{x}$ doesn't seem to work.
52766
How to solve the following integral $$\int_{0}^{\infty} \frac{ \ln x}{(x+a)(x+b)} dx,$$ where $a,b&gt;0$ and $a \neq b$. I was looking for some kind of substitution. However, I don't see an obvious one.  Thanks!
34440
I'm having a trouble with this integral expression:  $$\int_0^{2\pi} \frac{d\theta}{A+B \cos\theta}$$  I've done this substitution: $t= \tan(\theta/2)$  and get: $\displaystyle \cos\theta= \frac{1-t^2}{1+t^2}$ and $\displaystyle d\theta=\frac{2}{1+t^2}dt$ where $\displaystyle \cos^2\theta/2=\frac{1}{1+t^2}$  then the integral becomes:  $$\int\frac{2 \, dt}{(A-B)t^2+(A+B)}= \sqrt\frac{A+B}{A-B} \arctan \left(\left(\sqrt\frac{A+B}{A-B} \right) t\right)$$   However, I'm not sure about the new limits since $\tan$ has period $\pi$ so what I have to do at this point to decide the new limits? And of course if you find some mistake in what I've done before, please let me know!
63223
Given the integral $\int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx$  How can I evaluate it? Thanks!
45678
We had our final exam yesterday and one of the questions was to find out the value of: $$\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x $$ Interestingly enough, using the substitution $x=\frac{1}{t}$ we get - $$-\int_{0}^{1} \frac {\ln x}{1+x^2} \mathrm{d}x = \int_{1}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x $$and therefore $\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x = 0 $  I was curious to know about the theory behind this interesting (surprising even!) example.  Thank you.
19389
I have to compute this integral $$\int_{0}^{\infty} \frac{dt}{1+t^4}$$ to solve a problem in a homework. I have tried in many ways, but I'm stuck. A search in the web reveals me that it can be do it by methods of complex analysis. But I have not taken this course yet. Thanks for any help.
74654
I'm stuck in this problem. I can not integrate $\int \sec\theta\tan^2\theta \,d\theta$. Can anyone help me.
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.
104646
How would I carry out the following integration? $$\int_0^12\arctan x^2$$  I tried to substitute $x^2 = \tan\theta$ From there I wrote the integral as: $$\int_0^{\pi/4}\theta\cdot \frac{\sec^2\theta}{\sqrt{\tan\theta}}\ d\theta$$  Now, is my only option to use integration by parts? Or is there a better method to solve this integral?
30836
Calculate   $$\frac{\int_0^\pi x^3\ln(\sin x)\,dx}{\int_0^\pi x^2\ln(\sqrt{2}(\sin x))\,dx}$$   In this problem, I'm unable to understand how to start.  I tried applying integration by parts but I couldn't solve it. I also tried the various properties of definite integration but they were of no use. Maybe applying integration by parts (or DI method) successively may work but it leads to a form of $\frac{\infty}{\infty}$.
12916
How do I compute      $$\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$$   What I am doing is take   $$f(z)=\frac{(\log z)^2}{1+z^2}$$  and calculating   $\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}$ which came out to be $\dfrac{\pi}{2}-\dfrac{i\pi^2}{8}+\dfrac{i\pi}{2}$  Im not too sure how to move on from here. the given answer is $\dfrac{\pi^3}{8}$  Any help will be appreciated. thank you in advanced.
28772
How to solve the following integral.  $I(\theta) = \int_0^{\pi}\ln(1+\theta \cos x)dx$ where $|\theta|&lt;1$
114165
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$   $\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$  Put $\displaystyle x = \frac{1}{u},$ Then $\displaystyle dx = -\frac{1}{u^2}du$ and changing limits, We get  $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{1}{u(1+u^2)}du$$  So $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{t}{1+t^2}dt = \frac{1}{2}\int_{\frac{1}{e}}^{e}\frac{2t}{1+t^2}dt$$  So $$f(x) = \frac{1}{2}\ln (1+t^2)|_{\frac{1}{e}}^{e} = \ln(e) = 1$$  My question is how can we solve it in some shorter way, Help required, Thanks