Question

Solve:
${\int }_0^1{x}^3\sqrt{1+3x^4}dx$

Answer 1

$I={\int }_0^1{x}^3\sqrt{1+3x^4}dx$

$x^4=t\Rightarrow 4x^{3}dx=dt$

$\Rightarrow x^{3}dx=\frac{dt}{4}$

$\therefore I={\int }_0^1\sqrt{1+3t}\frac{dt}{4}$

$={\left[\frac{1}{\frac{3}{2}}\right(1+3t{)}^{\frac{3}{2}}.\frac{1}{4}]}_0^1$

$=\frac{1}{6}[4^{\frac{3}{2}}-1^{\frac{3}{2}}]$

$=\frac{8-1}{6}=\frac{7}{6}$