Answer
$$\frac{1}{2}x\sqrt {5{x^2} - 1} - \frac{{\sqrt 5 }}{2}{x^2}\ln \left| {\sqrt 5 x + \sqrt {5{x^2} - 1} } \right|$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {5{x^2} - 1} dx} \cr
& {\text{Rewrite}} \cr
& \int {\sqrt {5{x^2} - 1} dx} = \int {\sqrt {{{\left( {\sqrt 5 x} \right)}^2} - {{\left( 1 \right)}^2}} dx} \cr
& {\text{Let }}u = \sqrt 5 x,{\text{ and }}a = 1,{\text{ }}du = \sqrt 5 dx,{\text{ }}dx = \frac{1}{{\sqrt 5 }}du \cr
& {\text{Substitute}} \cr
& \int {\sqrt {{{\left( {\sqrt 5 x} \right)}^2} - {{\left( 1 \right)}^2}} dx} = \int {\sqrt {{u^2} - {a^2}} \left( {\frac{1}{{\sqrt 5 }}} \right)d} u \cr
& = \frac{1}{{\sqrt 5 }}\int {\sqrt {{u^2} - {a^2}} d} u \cr
& {\text{Use the special integration formula }}\left( {{\text{Theorem 8}}{\text{.2}}} \right) \cr
& \int {\sqrt {{u^2} - {a^2}} du = \frac{1}{2}\left( {u\sqrt {{u^2} - {a^2}} - {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right) + C} \cr
& \frac{1}{{\sqrt 5 }}\int {\sqrt {{u^2} - {a^2}} d} u = \frac{1}{{2\sqrt 5 }}\left( {u\sqrt {{u^2} - {a^2}} - {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right) + C \cr
& {\text{Substitute back}} \cr
& = \frac{1}{{2\sqrt 5 }}\left( {\sqrt 5 x\sqrt {{{\left( {\sqrt 5 x} \right)}^2} - {{\left( 1 \right)}^2}} - {{\left( {\sqrt 5 x} \right)}^2}\ln \left| {u + \sqrt {{{\left( {\sqrt 5 x} \right)}^2} - {{\left( 1 \right)}^2}} } \right|} \right) + C \cr
& = \frac{{\sqrt 5 x}}{{2\sqrt 5 }}\sqrt {5{x^2} - 1} - \frac{{5{x^2}}}{{2\sqrt 5 }}\ln \left| {\sqrt 5 x + \sqrt {5{x^2} - 1} } \right| \cr
& = \frac{1}{2}x\sqrt {5{x^2} - 1} - \frac{{\sqrt 5 }}{2}{x^2}\ln \left| {\sqrt 5 x + \sqrt {5{x^2} - 1} } \right| \cr} $$
