[tex]\int\limits_{0}^{ \frac{ \sqrt{2} }{2} } \frac{10x}{1 + 4x {}^{4} } \: dx \\ \\ \tg \theta = \frac{2x {}^{2} }{1} \: \to \: \tg \theta = 2x {}^{2} \: \to \: x = \sqrt{\frac{ \tg \theta}{2}} \\ \frac{dx}{d \theta} = \frac{1}{2} \left( \frac{ \tg \theta}{2} \right)^{ - \frac{1}{2} } . \frac{1}{2} \sec {}^{2} \theta \: \to \: dx = \frac{ \sec {}^{2} \theta }{4} . \left( \frac{ \tg \theta}{2} \right) {}^{ - \frac{1}{2} } d \theta \\ \\ \int\limits_{0}^{ \frac{ \sqrt{2} }{2} } \frac{10 \cancel{\sqrt{ \frac{ \tg \theta}{2} }} }{1 + 4. (\sqrt{ \frac{ \tg \theta}{2} )^{4} } }. \frac{ .\frac{ \sec {}^{2} \theta }{4} . \cancel{ \left( \frac{ \tg \theta}{2} \right) {}^{ - \frac{1}{2} }} d \theta}{1} \\ \\ \int\limits_{0}^{ \frac{ \sqrt{2} }{2} } \frac{ \frac{10 \sec {}^{2} \theta }{4} d \theta}{1 + 4 \frac{ \tg {}^{2} \theta}{4} } \: \to \int\limits_{0}^{ \frac{ \sqrt{2} }{2} } \frac{ \frac{5}{2} \sec {}^{2} \theta d \theta}{1 + \tg {}^{2} \theta \: } \\ \\ \boxed{1 + \tg {}^{2} \theta = \sec {}^{2} \theta} \\ \\ \int\limits_{0}^{ \frac{ \sqrt{2} }{2} } \frac{ \frac{5}{2} \sec {}^{2} \theta d \theta}{ \sec {}^{2} \theta } \: \to \: \int\limits_{0}^{ \frac{ \sqrt{2} }{2} } \frac{5}{2} d \theta \: \to \: \frac{5}{2} \left( \frac{ \theta}{1} \right)_{0}^{ \frac{ \sqrt{2} }{2} } \\ \\ \boxed{\tg \theta = 2x {}^{2} \: \to \: \theta = \arctan(2x {}^{2} )} \\ \\ \frac{5}{2} \left( \frac{ \arctan(2x {}^{2}) }{1} \right)_{0}^{ \frac{ \sqrt{2} }{2} } \\ \\ \frac{5}{2} . \arctan \left(2. \left( \frac{ \sqrt{2} }{2} \right) ^{2} \right) \: \to \: \frac{5}{2} . \arctan \left(1 \right) \: \to \: \frac{5}{2} . \frac{\pi}{4} \\ \\ \boxed{ \boxed{ \boxed{ \boxed{\frac{5\pi}{8} }}}}[/tex]