Tích phân hàm vô tỷ & siêu việt.

Đề bài: (Câu hỏi của bạn Nguyễn Hồng Sơn hỏi trên facebook Trợ Giúp Toán Học)
Tích tích phân: \[I = \int\limits_{\sqrt 3 }^{2\sqrt 2 } {\frac{{{x^3}\ln x}}{{\sqrt {{x^2} + 1} }}dx} \]Giải:
\[\begin{array}{l}
Do\,{I_1} = \int {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}dx = } \int {\frac{x}{{\sqrt {{x^2} + 1} }}.{x^2}.dx = } \\
 = \int {\left( {{{\left( {\sqrt {{x^2} + 1} } \right)}^2} - 1} \right)d\left( {\sqrt {{x^2} + 1} } \right)}  = \frac{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }}{3} - \sqrt {{x^2} + 1} \\
 \Rightarrow Coi\,\left\{ \begin{array}{l}
u = \ln x\\
dv = \frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \frac{{dx}}{x}\\
v = \frac{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }}{3} - \sqrt {{x^2} + 1}
\end{array} \right.\\
 \Rightarrow I = \left[ {\frac{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }}{3} - \sqrt {{x^2} + 1} } \right]\ln x\left| \begin{array}{l}
2\sqrt 2 \\
\sqrt 3
\end{array} \right. - \frac{1}{3}\int\limits_{\sqrt 3 }^{2\sqrt 2 } {\frac{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }}{x}dx}  + \int\limits_{\sqrt 3 }^{2\sqrt 2 } {\frac{{\sqrt {{x^2} + 1} }}{x}dx} \\
 = 9\ln 2 - \frac{1}{3}A + B\\
 \bullet A = \int\limits_{\sqrt 3 }^{2\sqrt 2 } {\frac{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }}{x}dx} .\,Coi\,t = \sqrt {{x^2} + 1}  \Leftrightarrow \left\{ \begin{array}{l}
{t^2} = {x^2} + 1 \Leftrightarrow 2tdt = 2xdx \Leftrightarrow dx = \frac{{tdt}}{x}\\
{x^2} = {t^2} - 1
\end{array} \right.\\
 \Rightarrow A = \int\limits_2^3 {\frac{{{t^4}}}{{{t^2} - 1}}du = } \int\limits_2^3 {\left( {{t^2} + 1 + \frac{1}{{{t^2} - 1}}} \right)du = \left( {\frac{{{t^3}}}{3} + t + \frac{1}{2}\ln \left| {\frac{{t - 1}}{{t + 1}}} \right|} \right)\left| \begin{array}{l}
3\\
2
\end{array} \right. = } \frac{{16}}{3} - \frac{1}{2}\ln \frac{2}{3}\\
 \bullet B = \int\limits_{\sqrt 3 }^{2\sqrt 2 } {\frac{{\sqrt {{x^2} + 1} }}{x}dx} .\,Coi\,t = \sqrt {{x^2} + 1}  \Leftrightarrow \left\{ \begin{array}{l}
{t^2} = {x^2} + 1 \Leftrightarrow 2tdt = 2xdx \Leftrightarrow dx = \frac{{tdt}}{x}\\
{x^2} = {t^2} - 1
\end{array} \right.\\
 \Rightarrow B = \int\limits_2^3 {\frac{{{t^2}}}{{{t^2} - 1}}du = } \int\limits_2^3 {\left( {1 + \frac{1}{{{t^2} - 1}}} \right)du = \left( {t + \frac{1}{2}\ln \left| {\frac{{t - 1}}{{t + 1}}} \right|} \right)\left| \begin{array}{l}
3\\
2
\end{array} \right. = } 1 - \frac{1}{2}\ln \frac{2}{3}\\
 \Rightarrow I = 9\ln 2 - \frac{1}{3}\left( {\frac{{16}}{3} - \frac{1}{2}\ln \frac{2}{3}} \right) + \left( {1 - \frac{1}{2}\ln \frac{2}{3}} \right) = 9\ln 2 - \frac{7}{9} - \frac{1}{3}\ln \frac{2}{3}
\end{array}\]