Question
Evaluate the following integrals:
$\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
$\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
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$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}$
Show that$\big[\text{G}(\beta)\big]^{-1}=\text{G}(-\beta)$
f : R → R, defined by f(x) = 1 + x2