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Câu hỏi: Tìm nguyên hàm của hàm số $f\left( x \right)={{x}^{3}}\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)$ ?
A. ${{x}^{4}}\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)-2{{x}^{2}}$.
B. $\left( \dfrac{{{x}^{4}}-16}{4} \right)\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)-2{{x}^{2}}$.
C. ${{x}^{4}}\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)+2{{x}^{2}}$.
D. $\left( \dfrac{{{x}^{4}}-16}{4} \right)\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)+2{{x}^{2}}$.
A. ${{x}^{4}}\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)-2{{x}^{2}}$.
B. $\left( \dfrac{{{x}^{4}}-16}{4} \right)\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)-2{{x}^{2}}$.
C. ${{x}^{4}}\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)+2{{x}^{2}}$.
D. $\left( \dfrac{{{x}^{4}}-16}{4} \right)\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)+2{{x}^{2}}$.
Đặt : $\left\{ \begin{aligned}
& u=\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right) \\
& dv={{x}^{3}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{16x}{{{x}^{4}}-16} \\
& v=\dfrac{{{x}^{4}}}{4}-4=\dfrac{{{x}^{4}}-16}{4} \\
\end{aligned} \right.$
$\Rightarrow \int{{{x}^{4}}\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)dx=\left( \dfrac{{{x}^{4}}-16}{4} \right)\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)-\int{4xdx}=\left( \dfrac{{{x}^{4}}-16}{4} \right)\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)-2{{x}^{2}}}+C$
Vậy đáp án đúng là đáp án B.
& u=\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right) \\
& dv={{x}^{3}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{16x}{{{x}^{4}}-16} \\
& v=\dfrac{{{x}^{4}}}{4}-4=\dfrac{{{x}^{4}}-16}{4} \\
\end{aligned} \right.$
$\Rightarrow \int{{{x}^{4}}\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)dx=\left( \dfrac{{{x}^{4}}-16}{4} \right)\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)-\int{4xdx}=\left( \dfrac{{{x}^{4}}-16}{4} \right)\ln \left( \dfrac{4-{{x}^{2}}}{4+{{x}^{2}}} \right)-2{{x}^{2}}}+C$
Vậy đáp án đúng là đáp án B.
Đáp án B.