Evaluate a definite integral via u-substitution

Let $u = x^4+1$. Then $du = 4x^3\,dx$, so $x^3\,dx = \frac{1}{4}du$. Changing the bounds: when $x=0$, $u=0^4+1=1$; when $x=1$, $u=1^4+1=2$. The integral becomes $\int_1^2 \sqrt{u}\cdot\frac{1}{4}\,du = \frac{1}{4}\int_1^2 u^{1/2}\,du.$

Now integrate: $\frac{1}{4}\int_1^2 u^{1/2}\,du = \frac{1}{4}\cdot\frac{2}{3}u^{3/2}\Big|_1^2 = \frac{1}{6}\left(2^{3/2} - 1^{3/2}\right)$. Since $2^{3/2} = 2\sqrt{2}$, this evaluates to $\frac{1}{6}(2\sqrt{2}-1)$, which is approximately $0.3047$.