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Câu hỏi: Tính diện tích hình phẳng giới hạn bởi các đường $y=\dfrac{2 x}{x+1} ; y=x^2 ; x=0 ; x=1$.
A. $\left|2 \ln 2-\dfrac{1}{3}\right|$
B. $\left|2 \ln 2-\dfrac{5}{3}\right|$
C. $\left|2 \ln 2-\dfrac{2}{3}\right|$
D. $\left|2 \ln 2-\dfrac{7}{3}\right|$
A. $\left|2 \ln 2-\dfrac{1}{3}\right|$
B. $\left|2 \ln 2-\dfrac{5}{3}\right|$
C. $\left|2 \ln 2-\dfrac{2}{3}\right|$
D. $\left|2 \ln 2-\dfrac{7}{3}\right|$
Diện tích hình phẳng giới hạn bởi các đường $y=\dfrac{2 x}{x+1} ; y=x^2 ; x=0 ; x=1$ là:
$
\begin{gathered}
S=\int_0^1\left|x^2-\dfrac{2 x}{x+1}\right| \mathrm{dx}=\left|\int_0^1\left(x^2-\dfrac{2 x}{x+1}\right) \mathrm{dx}\right|=\left|\int_0^1\left(x^2-2+\dfrac{2}{x+1}\right) \mathrm{dx}\right| \\
=\left|\left(\dfrac{1}{3} x^3-2 x+2 \ln |x+1|\right)\right|_0^1 \mid \\
=\left|2 \ln 2-\dfrac{5}{3}\right| .
\end{gathered}
$
$
\begin{gathered}
S=\int_0^1\left|x^2-\dfrac{2 x}{x+1}\right| \mathrm{dx}=\left|\int_0^1\left(x^2-\dfrac{2 x}{x+1}\right) \mathrm{dx}\right|=\left|\int_0^1\left(x^2-2+\dfrac{2}{x+1}\right) \mathrm{dx}\right| \\
=\left|\left(\dfrac{1}{3} x^3-2 x+2 \ln |x+1|\right)\right|_0^1 \mid \\
=\left|2 \ln 2-\dfrac{5}{3}\right| .
\end{gathered}
$
Đáp án B.