Answer
$$\frac{1}{5}\ln \left( {5 + \sqrt {26} } \right)$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{1}{{\sqrt {25{x^2} + 1} }}} dx \cr
& {\text{Let }}u = 5x,{\text{ }}du = 5dx,{\text{ }}du = \frac{1}{5}du \cr
& {\text{The new limitts of integration are}} \cr
& x = 1 \to u = 5 \cr
& x = 0 \to u = 0 \cr
& {\text{Substituting}} \cr
& \int_0^1 {\frac{1}{{\sqrt {25{x^2} + 1} }}} dx = \int_0^5 {\frac{1}{{\sqrt {{u^2} + 1} }}\left( {\frac{1}{5}} \right)du} \cr
& {\text{Use Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{\sqrt {{u^2} + 1} }} = } \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr
& \int_0^5 {\frac{1}{{\sqrt {{u^2} + 1} }}\left( {\frac{1}{5}} \right)du} = \frac{1}{5}\left[ {\ln \left( {u + \sqrt {{u^2} + 1} } \right)} \right]_0^5 \cr
& = \frac{1}{5}\left[ {\ln \left( {5 + \sqrt {{5^2} + 1} } \right)} \right] - \frac{1}{5}\left[ {\ln \left( {0 + \sqrt {{0^2} + 1} } \right)} \right] \cr
& = \frac{1}{5}\left[ {\ln \left( {5 + \sqrt {26} } \right)} \right] - \frac{1}{5}\left[ {\ln \left( 1 \right)} \right] \cr
& = \frac{1}{5}\ln \left( {5 + \sqrt {26} } \right) \cr} $$
