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Câu hỏi: Tích phân $I=\int\limits_{-1}^{1}{\dfrac{x}{\sqrt{x+1}-1}dx}$ có giá trị là:
A. $I=\dfrac{4\sqrt{2}}{3}+2$
B. $I=\dfrac{4\sqrt{2}}{3}-2$
C. $I=\dfrac{4\sqrt{2}}{3}-1$
D. $I=\dfrac{4\sqrt{2}}{3}+1$
A. $I=\dfrac{4\sqrt{2}}{3}+2$
B. $I=\dfrac{4\sqrt{2}}{3}-2$
C. $I=\dfrac{4\sqrt{2}}{3}-1$
D. $I=\dfrac{4\sqrt{2}}{3}+1$
$\begin{aligned}
& \dfrac{x}{\sqrt{x+1}-1}=\sqrt{x+1}+1\Rightarrow I=\int\limits_{-1}^{1}{\dfrac{x}{\sqrt{x+1}-1}}dx=\int\limits_{-1}^{1}{\left( \sqrt{x+1}+1 \right)}dx \\
& =\left. \left[ \dfrac{2}{3}{{\left( x+1 \right)}^{\dfrac{3}{2}}}+x \right] \right|_{-1}^{1}=\dfrac{4\sqrt{2}}{3}+2 \\
\end{aligned}$.
& \dfrac{x}{\sqrt{x+1}-1}=\sqrt{x+1}+1\Rightarrow I=\int\limits_{-1}^{1}{\dfrac{x}{\sqrt{x+1}-1}}dx=\int\limits_{-1}^{1}{\left( \sqrt{x+1}+1 \right)}dx \\
& =\left. \left[ \dfrac{2}{3}{{\left( x+1 \right)}^{\dfrac{3}{2}}}+x \right] \right|_{-1}^{1}=\dfrac{4\sqrt{2}}{3}+2 \\
\end{aligned}$.
Đáp án A.