ลองทำนะครับ...
ให้ $\sqrt[4]{x} = \tan u $
$$\begin{array}{rcl}\int \frac{\tan u}{1+\tan^2u}\ d(\tan^4u)&=&\int\frac{\tan u}{ \sec^2u} d(\tan^4 u)\\
\because\quad d(\tan^4u)&=& 4\tan^3u\sec^2u \ du\\
ดังนั้น \quad \int \frac{\tan u}{\sec^2u}\ d(\tan^4u)&=&\int 4\tan^4u\ d u\\
\because\quad \int \tan^4u\ du&=&\int (\sec^2u-1)\tan^2u \ du \\
&=&\int\sec^2u\tan^2u\ du-\int\tan^2u \ du \\
&=&\int\tan^2u\ d(\tan u)-\int\sec^2u\ du+\int \ du\\
&=&\frac{\tan^3u}{3}-\tan u +u+C\\
\int 4\tan^4u\ du&=&4(\frac{\tan^3u}{3}-\tan u +u+C)\\
\because \ u&=&\arctan\sqrt[4]{x}\\
\int \frac{ \sqrt[4]{x}}{1+\sqrt x}&=&4\big(\frac{x^{3/4}}{3}-\sqrt[4]{x}+\arctan{\sqrt[4]{x}}\big)\\
\therefore \int_0^{16} \frac{ \sqrt[4]{x}}{1+\sqrt x}&=&4(\frac{8}{3}-2+\arctan 2)\\
&=&4(\frac{2}{3}+\arctan 2 )
\end{array}$$
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Edit: แก้ขอบเขตครับ