Hello, 3deltat!
$\displaystyle 3)\;\;\int^{\sqrt{3}}_1 \arctan\left(\frac{1}{x}\right)\,dx$
I got the integral to come out as: .$\displaystyle x\arctan\left(\frac{1}{x}\right) + \frac{1}{2}\ln(x^2 +1) $ . . . . Right!
However, this is wrong, because i know that the answer SHOULD be 0.468 . ?? How can the answer be negative? . . . The graph is above the xaxis. Code:

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Ah, I see skeeter already beat me to it!
We have: .$\displaystyle x\arctan\left(\frac{1}{x}\right) + \frac{1}{2}\ln(x^2+1)\:\bigg]^{\sqrt{3}}_1$
. . $\displaystyle \bigg[\sqrt{3}\arctan\left(\frac{1}{\sqrt{3}}\right) + \frac{1}{2}\ln(4)\bigg]  \bigg[1\!\cdot\!\arctan(1) + \frac{1}{2}\ln(2)\bigg] $
. . $\displaystyle = \;\bigg[\sqrt{3}\left(\frac{\pi}{6}\right) + \ln\left(4^{\frac{1}{2}}\right)\bigg]  \bigg[\frac{\pi}{4} + \frac{1}{2}\ln(2)\bigg] $
. . $\displaystyle = \;\frac{\pi\sqrt{3}}{6} + \ln(2)  \frac{\pi}{4}  \frac{1}{2}\ln(2) $
. . $\displaystyle = \;\left(\frac{2\sqrt{3}3}{12}\right)\pi + \frac{1}{2}\ln(2) $
. . $\displaystyle = \;\boxed{0.468}075109$