Evaluate $\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2 } \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $ 
Evaluate
   $$\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2
 } \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $$

The question look's like there is a nice method to do it, but I can't figure out. Can someone provide some hint or answer?.
 A: Set R = 4 x3 + 5 x2 + 10 for brevity. Maybe based on Liouville's theorem, we can guess
$$\int \sqrt{R} \left( x^5 + x^4 + x^2 \right) = PR^{3/2} + c$$
where P is a polynomial of x and c is the constant of integration. Differentiate on both sides, we get
$$\sqrt{R} \left( x^5 + x^4 + x^2 \right) = P'R^{3/2} + PR' \sqrt{R}.$$
Rationalize the formula
$$\left( x^5 + x^4 + x^2 \right) = \left( 4x^3 + 5x^2 + 10 \right) P' + \left(
12x^2 + 10x \right) P.$$
P is cubic, so set P = a3x3 + a2x2 + a1x + a0 and solve the linear system. The system is overdeterined, but luckily there is still a solution
$$P = \frac{x^3}{30}.$$
As a result,
$$\int \sqrt{4x^3 + 5x^2 + 10} \left( x^5 + x^4 + x^2 \right) = \frac{x^3 \left( 4x^3 + 5x^2 + 10 \right)^{3/2}}{30} + c.$$
Therefore
$$\int_0^1 \sqrt{4x^3 + 5x^2 + 10} \left( x^5 + x^4 + x^2 \right) = \frac{19^{3/2}}{30}.$$
A: $\displaystyle{\large%
\int_{0}^{1}
\left(x^{5} + x^{4} + x^{2}\right)\,
\sqrt {4x^{3} + 5x^2 + 10\;}\;{\rm d}x\quad:{\Huge ?}}$
Following @jdh8 ( $\color{#0000ff}{\mbox{There is a missing}\; \color{#ff0000}{\large 3/2}\
\mbox{factor in his formula}}$ ).
$P \equiv a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}$
\begin{align}
\left(x^{5} + x^{4} + x^{2}\right)
&=
\left(4x^{3} + 5x^{2} + 10 \right)P'
+ 
\left(18x^{2} + 15x\right)P 
\\[3mm]
{x^{4} + x^{3} + x \over 18x + 15}
&=
{4x^{3} + 5x^{2} + 10 \over 18x^{2} + 15x}\;P'
+
P 
\end{align}
Take the limit $x \to 0$. In order to save a divergence we'll get $a_{1} = 0$ and
it follows that $a_{0} = 0$. Then
$$
{x^{4} + x^{3} + x \over 18x + 15}
=
{4x^{3} + 5x^{2} + 10 \over 18x + 15}\;\left(3a_{3} x + 2a_{2}\right)
+
\left[a_{3}x^{3} + a_{2}x^{2}\right] 
$$
Again, take the limit $x \to 0$ and we get $a_{2} = 0$. The last expression is reduced to
$$
{x^{3} + x^{2} + 1 \over 18x + 15}
=
{4x^{3} + 5x^{2} + 10 \over 18x + 15}\;\left(3a_{3}\right)
+
\left[a_{3}x^{2}\right] 
$$
One more time
$$
x \to 0
\quad\Longrightarrow\quad
{1 \over 15} = {10 \over 15}\left(3a_{3}\right)
\quad\Longrightarrow\quad
a_{3} = {1 \over 30}
\quad\Longrightarrow\quad
\color{#ff0000}{\large P = {x^{3} \over 30}}
$$
A: $$\begin{align}
\int_0^1(x^5+x^4+x^2)\sqrt{4x^3+5x^2+10}dx &=\int_0^1(x^4+x^3+x)\sqrt{4x^5+5x^4+10x^2}dx\cr
&={1\over20}\int_0^{19}\sqrt{u}du={19^{3/2}\over30}\cr
\end{align}$$
