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Câu hỏi: Cho nguyên hàm của $\int{{{x}^{2}}\ln x}\text{d}x=a{{x}^{3}}\ln x-b{{x}^{3}}+C$ trong đó $a,b,c\in \mathbb{R}$. Tính giá trị $T=a+b$
A. $T=\dfrac{4}{9}$.
B. $T=\dfrac{5}{9}$.
C. $T=\dfrac{2}{9}$.
D. $T=\dfrac{1}{3}$.
A. $T=\dfrac{4}{9}$.
B. $T=\dfrac{5}{9}$.
C. $T=\dfrac{2}{9}$.
D. $T=\dfrac{1}{3}$.
$\int{{{x}^{2}}\ln x\text{d}x}$
Đặt $\left\{ \begin{aligned}
& {{x}^{2}}\text{d}x=\text{d}v \\
& \ln x=u \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& v=\dfrac{1}{3}{{x}^{3}} \\
& \text{d}u=\dfrac{1}{x}\text{d}x \\
\end{aligned} \right.$.
Suy ra $\int{{{x}^{2}}\ln x\text{d}x}=\dfrac{1}{3}{{x}^{3}}\ln x-\int{\dfrac{1}{3}{{x}^{2}}\text{d}x=}\dfrac{{{x}^{3}}}{3}\ln x-\dfrac{1}{9}{{x}^{3}}+C.$
$T=a+b=\dfrac{1}{3}+\dfrac{1}{9}=\dfrac{4}{9}$.
Đặt $\left\{ \begin{aligned}
& {{x}^{2}}\text{d}x=\text{d}v \\
& \ln x=u \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& v=\dfrac{1}{3}{{x}^{3}} \\
& \text{d}u=\dfrac{1}{x}\text{d}x \\
\end{aligned} \right.$.
Suy ra $\int{{{x}^{2}}\ln x\text{d}x}=\dfrac{1}{3}{{x}^{3}}\ln x-\int{\dfrac{1}{3}{{x}^{2}}\text{d}x=}\dfrac{{{x}^{3}}}{3}\ln x-\dfrac{1}{9}{{x}^{3}}+C.$
$T=a+b=\dfrac{1}{3}+\dfrac{1}{9}=\dfrac{4}{9}$.
Đáp án A.