Find the value of following function's integral  with respect to x from 0 to 1  $$F(x)=\frac{\ln(1+x)}{1+x^2}$$
122105
Possible Duplicate:          $$\int_{0}^{1} dx\frac{\log(1+x)}{1 + x^2}$$  I am having a hard time deriving the answer, $\frac{\pi}{8} \log(2) $.  I have tried Taylor expansion of both numerator and denominator, both seem too complicated and fruitless.
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 $\frac{\infty}{\infty}$.
107906
I have to verify that     $$\int_0^\pi \ln(1+\alpha\cos(x))dx=\pi\ln\left(\frac{1+\sqrt{1-\alpha^2}}{2}\right)$$   with $|\alpha|&lt;1$. It is my homework and don't know where to begin.
115119
Does anyone know the exact value of this integral?   $$\int^1_0 \frac{p^2-1}{\ln p}dp$$   I know that the approximate value is $1.09861...$. But I can't seem to get a figure on the exact value, or an antiderivative.
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?
78427
I'm trying to find the value of the integral $\int_{-4}^{-2}e^{-x}\,dx$ but I just couldn't solve it.  Actually I found in a  that $\int e^x\,dx=e^x+C$ so I concluded:  $$ \int e^{-x}\,dx=\int\frac{1}{e^x}\,dx=\ln|e^x|$$  $$\int\limits_{-4}^{-2}e^{-x}\,dx=\left(\ln|e^{-2}|\right)-\left(\ln|e^{-4}|\right)=-2+4=2$$  I know the solution is wrong, but how can I solve this integral or any another integral like 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.
20215
Question says it all, how can I verify the following?  $$\int\sec x\ dx=\frac12 \ln \left|\frac{1+\sin x}{1-\sin x}\right| + C$$
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
66722
Find the limit of function  $$\lim_{x\to 0}\left(\frac{(1+x)^\frac1x}{e}\right)^\frac1x$$
92294
Calculate the integral $$\int \ln (\sin x) \, dx.$$
60941
How do I integrate the following function: $$\int_{0}^{\infty} \frac{\log{x}}{x^{2}+a^{2}} \ dx $$ Integrating by parts looks very difficult.
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.
89061
I want to compute the following integral: $$\raise 1ex{\Large\int} \frac{\sqrt{\ln(x+\sqrt{1+x^2}})}{1+x^2}\,dx$$
66782
I'm trying to find $$ \lim_{n \to \infty}   \int_1^a \dfrac{n}{1+x^n} \, dx  ,$$ where parameter $a&gt;1$. The limit appears to be $\ln 2$ regardless of the value of $a$. Some trick is probably needed here, but i don't see it.
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,
39323
I am trying to evaluate  $$ \int_0^1 \frac{\ln(1+x)}{x}\,dx $$  I started by using the Taylor series for $\ln (1+x)$  $$\begin{align*} \int_0^1 \frac{\ln(1+x)}{x}\,dx &amp;= \int_0^1\frac{1}{x}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}dx \\&amp;=\int_0^1 \frac{1}{x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+...\right)dx \\&amp;=\int_0^1 \left(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\frac{x^4}{5}-\frac{x^5}{6}+...\right)dx\\&amp;=1-\frac{1}{4}+\frac{1}{9} -\frac{1}{16}+\frac{1}{25}-\frac{1}{36}+... \\ &amp;=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\end{align*} $$  I'm aware of the fact that $$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$$ however the $(-1)^{n+1}$ is giving me trouble.