1+x⁴=t, derivam, 4t³dx=dt
x=0, t=1; x=1, t=2
integrala devine
¹²[tex] \frac{1}{4} \int\limits^2_1 {4x \sqrt{t} } \, dx = \frac{1}{4} \int\limits^2_1 { \sqrt{t} } \, dt = \\ \frac{1}{4} \int\limits^2_1 { t^{ \frac{1}{2} } } \, dt = \frac{1}{4} \frac{1}{ \frac{1}{2} +1} t^{ \frac{3}{2} } = \\ \frac{1}{4} \frac{2}{3} t^{ \frac{3}{2} } = \frac{1}{6} (2 \sqrt{2} -1)[/tex]
