## Calculus (3rd Edition)

$$\frac{1}{10}\ln\frac{3}{2}=0.0405.$$
Using the method of partial fractions, we have $$\int_0^1\frac{dx}{25-x^2}=\frac{1}{10}\int_0^1 \frac{1}{5-x}+\frac{1}{5+x}dx \\ =\frac{1}{10}(-\ln(5-x)+\ln(5+x))|_0^1=\frac{1}{10}(-\ln4+\ln6+\ln5-\ln5)\\ =\frac{1}{10} \ln \frac{6}{4}=\frac{1}{10}\ln\frac{3}{2}=0.0405.$$