## Calculus (3rd Edition)

$4.95 \times 10^{-5}.$
The total charge is given by $$\int_{0}^{10} \rho(x) d x=10^{-4} \int_{0}^{10} \frac{x}{\left(x^{2}+1\right)^{2}} d x\\ =\left.10^{-4}\left(-\frac{1}{2}\left(x^{2}+1\right)^{-1}\right)\right|_{0} ^{10}\\ =5 \times 10^{-5}\left(1-\frac{1}{101}\right)=4.95 \times 10^{-5}.$$