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

$$\begin{align} \int_0^1(x^5+x^4+x^2)\sqrt{4x^3+5x^2+10}dx &=\int_0^1(x^4+x^3+x)\sqrt{4x^5+5x^4+10x^2}dx\cr &={1\over20}\int_0^{19}\sqrt{u}du={19^{3/2}\over30}\cr \end{align}$$

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 $$\int_0^1 \sqrt{4x^3 + 5x^2 + 10} \left( x^5 + x^4 + x^2 \right) = \frac{19^{3/2}}{30}.$$

$\displaystyle{\large% \int_{0}^{1} \left(x^{5} + x^{4} + x^{2}\right)\, \sqrt {4x^{3} + 5x^2 + 10\;}\;{\rm d}x\quad:{\Huge ?}}$

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 $$ x \to 0 \quad\Longrightarrow\quad {1 \over 15} = {10 \over 15}\left(3a_{3}\right) \quad\Longrightarrow\quad a_{3} = {1 \over 30} \quad\Longrightarrow\quad \color{#ff0000}{\large P = {x^{3} \over 30}} $$