Đáp án:
\[\frac{1}{2}\left[ {{x^2}.\ln \left( {{x^2} + 1} \right) - {x^2} + \ln \left( {{x^2} + 1} \right)} \right]\]
Giải thích các bước giải:
Đặt:
\(\begin{array}{l}
\left\{ \begin{array}{l}
u = \ln \left( {{x^2} + 1} \right)\\
v' = x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \frac{{2x}}{{{x^2} + 1}}\\
v = \frac{{{x^2}}}{2}
\end{array} \right.\\
\int {x.\ln \left( {{x^2} + 1} \right)dx} = \frac{{{x^2}}}{2}.\ln \left( {{x^2} + 1} \right) - \int {\frac{{2x}}{{{x^2} + 1}}.\frac{{{x^2}}}{2}dx} \\
= \frac{{{x^2}}}{2}.\ln \left( {{x^2} + 1} \right) - \int {\frac{{{x^3}}}{{{x^2} + 1}}dx} \\
= \frac{{{x^2}}}{2}.\ln \left( {{x^2} + 1} \right) - \int {\left( {x - \frac{x}{{{x^2} + 1}}} \right)dx} \\
= \frac{{{x^2}}}{2}\ln \left( {{x^2} + 1} \right) - \frac{{{x^2}}}{2} + \int {\frac{{xdx}}{{{x^2} + 1}}} \\
= \frac{{{x^2}}}{2}.\ln \left( {{x^2} + 1} \right) - \frac{{{x^2}}}{2} + \int {\frac{{\frac{1}{2}d\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}} \\
= \frac{{{x^2}}}{2}\ln \left( {{x^2} + 1} \right) - \frac{{{x^2}}}{2} + \frac{1}{2}.ln\left( {{x^2} + 1} \right)\\
= \frac{1}{2}\left[ {{x^2}.\ln \left( {{x^2} + 1} \right) - {x^2} + \ln \left( {{x^2} + 1} \right)} \right]
\end{array}\)
