Tính $\ I=\int\limits_{\frac{1}{3}}^{1}{\left( \frac{1}{3x+2+2\sqrt{3x+1}}+\frac{\sqrt[3]{x-{{x}^{3}}}}{{{x}^{6}}} \right)dx}.$

Đề bài: (Câu hỏi của bạn Micale Cat trên facebook Trợ Giúp Toán Học)
Tính tích phân: $\ I=\int\limits_{\frac{1}{3}}^{1}{\left( \frac{1}{3x+2+2\sqrt{3x+1}}+\frac{\sqrt[3]{x-{{x}^{3}}}}{{{x}^{6}}} \right)dx}.$
Giải:
Đặt:$\ {{I}_{1}}=\int\limits_{\frac{1}{3}}^{1}{\frac{dx}{3x+2+2\sqrt{3x+1}}};\,{{I}_{2}}=\int\limits_{\frac{1}{3}}^{1}{\frac{\sqrt[3]{x-{{x}^{3}}}}{{{x}^{6}}}dx}.$
+ Xét $\ {{I}_{1}}=\int\limits_{\frac{1}{3}}^{1}{\frac{dx}{3x+2+2\sqrt{3x+1}}}=\int\limits_{\frac{1}{3}}^{1}{\frac{dx}{3x+1+2\sqrt{3x+1}+1}}=\int\limits_{\frac{1}{3}}^{1}{\frac{dx}{{{\left( \sqrt{3x+1}+1 \right)}^{2}}}}.$
 Đặt $\ t=\sqrt{3x+1}\Leftrightarrow {{t}^{2}}=3x+1\Rightarrow dx=\frac{2}{3}tdt$ $\ \Rightarrow {{I}_{1}}=\frac{2}{3}\int\limits_{\sqrt{2}}^{2}{\frac{tdt}{{{\left( t+1 \right)}^{2}}}}=\frac{2}{3}\int\limits_{\sqrt{2}}^{2}{\left( \frac{1}{t+1}-\frac{1}{{{\left( t+1 \right)}^{2}}} \right)d\left( t+1 \right)}.$

$\ =\frac{2}{3}\left[ \ln \left| t+1 \right|+\frac{1}{t+1} \right]\left| \begin{align}
  & 2 \\
 & \sqrt{2} \\
\end{align} \right.=\frac{2}{3}\ln 3\left( \sqrt{2}-1 \right)+\frac{8-6\sqrt{2}}{9}.$
+ Xét $\ {{I}_{2}}=\int\limits_{\frac{1}{3}}^{1}{\frac{\sqrt[3]{x-{{x}^{3}}}}{{{x}^{6}}}dx}=\int\limits_{\frac{1}{3}}^{1}{\frac{1}{{{x}^{5}}}\sqrt[3]{\frac{1}{{{x}^{2}}}-1}\,dx}.$
Đặt $\ u=\sqrt[3]{\frac{1}{{{x}^{2}}}-1}\Leftrightarrow {{u}^{3}}=\frac{1}{{{x}^{2}}}-1\Rightarrow 3{{u}^{2}}du=-\frac{2}{{{x}^{3}}}dx\Leftrightarrow dx=-\frac{3{{u}^{2}}.{{x}^{3}}du}{2}.$
$\ \Rightarrow {{I}_{2}}=\frac{3}{2}\int\limits_{0}^{2}{\left( {{u}^{3}}+1 \right){{u}^{3}}du=\frac{3}{2}}\left( \frac{{{u}^{7}}}{7}+\frac{{{u}^{4}}}{4} \right)\left| \begin{align}
  & 2 \\
 & 0 \\
\end{align} \right.=\frac{234}{7}\Rightarrow I={{I}_{1}}+{{I}_{2}}=\frac{2}{3}\ln 3\left( \sqrt{2}-1 \right)+\frac{2162-42\sqrt{2}}{63}.$