So one of the question on the MIT Integration bee has baffled me all day today $$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$$ I have tried a variety of things to do this, starting with Integration By Parts Part 1 $$\frac{\tan x-x}{1+x^2}\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac...

Show that $$\int_{-\infty}^\infty \frac{\log\left(\sqrt{x^2+a^2}\right)}{x^2+b^2}\,dx = \frac{\pi}{b}\log\left(a+b\right)$$ for $a,b>0\in\mathbb{R}$. I stumbled on this answer empirically, but I'm not sure how to solve it directly.

The general formula (for $m > n+1$ and $n \ge 0$) is $\frac{\pi}{m} \csc\left(\frac{\pi (n+1)}{m}\right)$. IIRC the usual method involves a wedge-shaped contour of angle $2 \pi/m$. EDIT: Consider $\oint_\Gamma f(z)\ dz$ where $f(z) = \frac{z^n}{1+z^m}$ (using the principal branch if $m$ or $n$ ...

Context: If you guys didn't know, I'm running a nice little contest to see who can program the largest number. More specific rules if you are interested may be found in my chat room (click here to join). If you are entering, do note that I am accepting entries for quite a while (ignore all the ...

Using synthetic division, we have that . $$\begin{array}{c|rrr} & 2 & 0 & -19 & 3\\ {\color{red}3} & \downarrow & 6 & 18 & -3\\ \hline & 2 & 6 & -1 & |\phantom{-} {\color{blue}0} \end{array}$$ Thus, $$2x^3-19x+3=(x-3)(2x^2+6x-1)$$