Integration Fun

Simpler solution from @John Frank .

$\begin{aligned} I & = \int_{\sqrt[4]{25}}^{\sqrt[4]{36}} \frac {16x^3-8x}{x^4-x^2} dx & \small \color{#3D99F6} \text{Let }u = x^4-x^2 \implies du = 4x^3-2x \ dx \\ & = \int_{20}^{30} \frac {4du}{u} \\ & = 4 \ln u \ \bigg|_{20}^{30} \\ & = 4 \ln \frac 32 \\ & = \ln \frac {81}{16} \end{aligned}$

Therefore, $\sqrt{\dfrac {81}{16}} = \dfrac 94 = 2.25$ and its digital sum is $2+2+5=\boxed{9}$ .

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$\begin{aligned} I & = \int_{\sqrt[4]{25}}^{\sqrt[4]{36}} \frac {16x^3-8x}{x^4-x^2} dx & \small \color{#3D99F6} \text{Simplify} \\ & = 8 \int_{\sqrt 5}^{\sqrt 6} \frac {2x^2-1}{x(x-1)(x+1)} dx & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = 4 \int_{\sqrt 5}^{\sqrt 6} \left(\frac 2x + \frac 1{x-1} + \frac 1{x+2} \right) dx \\ & = 4 \bigg[2\ln x + \ln (x-1) + \ln (x+1)\bigg]_{\sqrt 5}^{\sqrt 6} \\ & = 4 \bigg[\ln (x^4-x^2) \bigg]_{\sqrt 5}^{\sqrt 6} \\ & = 4 \left(\ln (36-6) - \ln(25-5) \right) \\ & = 4 \ln \frac 32 \\ & = \ln \frac {81}{16} \end{aligned}$

Therefore, $\sqrt {\dfrac {81}{16}} = \dfrac 94 = 2.25$ and its sum of digits is $2+2+5=\boxed{9}$ .

Factor out 8x in the numerator and $x^2$ in the bottom $\displaystyle \int_{\sqrt[4]{36} }^{\sqrt[4]{25} } \frac{8x(x^2-1)}{x^2(x^2-1)}$ = $\displaystyle \int_{\sqrt[4]{36} }^{\sqrt[4]{25} } \frac{8}{x}$ = $8\ln|x| = \ln(x^8)$ Evaluate from $\sqrt[4]{36}$ to $\sqrt[4]{25}$

$\displaystyle\ln36^\frac{8}{4}-\ln25^\frac{8}{4}$ = $\displaystyle\ln\frac{36^2}{25^2}$

Take square root $\displaystyle\sqrt{\frac{a}{b}}$

$\displaystyle\sqrt{\frac{36^2}{25^2}}$ = $\displaystyle\sqrt{(\frac{36}{25})^2}$ = $\displaystyle|\frac{36}{25}|$ = 1.44; 1+4+4 = $\boxed{9}$