$\displaystyle\int_{-1}^{1} 5x^4\sqrt{x^5+1}\,dx$ equals

1. Put $t=x^5+1$, so $dt=5x^4\,dx$. 2. Change limits: $x=-1\Rightarrow t=0$; $x=1\Rightarrow t=2$. 3. Integral $=\int_0^2 \sqrt{t}\,dt=\dfrac{2}{3}\left[t^{3/2}\right]_0^2=\dfrac{2}{3}\cdot 2\sqrt{2}$. 4. $=\dfrac{4\sqrt{2}}{3}$. _Source: syllabus Mathematics Ch 7 "Integrals", p.340_