Answer
$$\frac{5}{4}\sec 2{x^2} + C$$
Work Step by Step
$$\eqalign{
& \int {5x} \sec 2{x^2}\tan 2{x^2}dx \cr
& {\text{set }}u = 2{x^2}{\text{ then }}\frac{{du}}{{dx}} = 4x,\,\,\,\,\,\,\,\,\frac{{du}}{{4x}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {5x} \sec 2{x^2}\tan 2{x^2}dx = \int {5x\sec u\tan u} \left( {\frac{{du}}{{4x}}} \right) \cr
& {\text{cancel the common factor }}{x^4} \cr
& = \int {5\sec u\tan u} \left( {\frac{{du}}{4}} \right) \cr
& = \frac{5}{4}\int {\sec u\tan u} du \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {\sec x\tan x} dx = \sec x + C \cr
& = \frac{5}{4}\sec u + C \cr
& {\text{write in terms of }}x \cr
& = \frac{5}{4}\sec 2{x^2} + C \cr} $$
