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Câu hỏi: Biết $I=\int_{0}^{4}{x\ln \left( 2x+1 \right)}dx=\dfrac{a}{b}\ln 3-c$ với $a,b,c$ là các số nguyên và $\dfrac{a}{b}$ là phân số tối giản. Tính $T=a+b+c$.
A. $T=64$.
B. $T=68$.
C. $T=60$.
D. $T=70$
A. $T=64$.
B. $T=68$.
C. $T=60$.
D. $T=70$
Gọi $z=x+yi; w=a+bi$
Đặt $\left\{ \begin{aligned}
& u=\ln \left( 2x+1 \right) \\
& dv=xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{2}{2x+1} \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
$I=\left. \dfrac{{{x}^{2}}}{2}.\ln \left( 2x+1 \right) \right|_{0}^{4}-\int_{0}^{4}{\dfrac{{{x}^{2}}}{2x+1}dx}$
$=16\ln 3-\int_{0}^{4}{\left( \dfrac{x}{2}-\dfrac{1}{4}+\dfrac{1}{4\left( 2x+1 \right)} \right)dx}$ $=16\ln 3-\left. \left( \dfrac{{{x}^{2}}}{4}-\dfrac{x}{4}+\dfrac{1}{8}\ln \left| 8x+4 \right| \right) \right|_{0}^{4}$
$$ $=16\ln 3-\left( 3+\dfrac{1}{4}\ln 6-\dfrac{1}{4}\ln 2 \right)$ $=16\ln 3-3-\dfrac{1}{4}\ln 3=\dfrac{63}{4}\ln 3-3$
Vậy $a+b+c=63+4+3=70$
Đặt $\left\{ \begin{aligned}
& u=\ln \left( 2x+1 \right) \\
& dv=xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{2}{2x+1} \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
$I=\left. \dfrac{{{x}^{2}}}{2}.\ln \left( 2x+1 \right) \right|_{0}^{4}-\int_{0}^{4}{\dfrac{{{x}^{2}}}{2x+1}dx}$
$=16\ln 3-\int_{0}^{4}{\left( \dfrac{x}{2}-\dfrac{1}{4}+\dfrac{1}{4\left( 2x+1 \right)} \right)dx}$ $=16\ln 3-\left. \left( \dfrac{{{x}^{2}}}{4}-\dfrac{x}{4}+\dfrac{1}{8}\ln \left| 8x+4 \right| \right) \right|_{0}^{4}$
$$ $=16\ln 3-\left( 3+\dfrac{1}{4}\ln 6-\dfrac{1}{4}\ln 2 \right)$ $=16\ln 3-3-\dfrac{1}{4}\ln 3=\dfrac{63}{4}\ln 3-3$
Vậy $a+b+c=63+4+3=70$
Đáp án D.