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Câu hỏi: Có bao nhiêu số thực $a$ để $\int\limits_{0}^{1}{\dfrac{x}{a+{{x}^{2}}}dx}=1?$
A. 0.
B. 1.
C. 2.
D. 3.
A. 0.
B. 1.
C. 2.
D. 3.
$a+{{x}^{2}}\ne 0$ với mọi $x\in \left[ 0;1 \right]\Rightarrow a>0$ hoặc $a<-1.$
$\int\limits_{0}^{1}{\dfrac{x}{a+{{x}^{2}}}dx}=1\Leftrightarrow \dfrac{1}{2}\ln \left| a+{{x}^{2}} \right|\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{1}{2}\ln \left| \dfrac{a+1}{a} \right|=1\Leftrightarrow \left[ \begin{aligned}
& a=\dfrac{1}{{{e}^{2}}-1} \\
& a=-\dfrac{1}{{{e}^{2}}+1}\left( lọai \right) \\
\end{aligned} \right.$
$\int\limits_{0}^{1}{\dfrac{x}{a+{{x}^{2}}}dx}=1\Leftrightarrow \dfrac{1}{2}\ln \left| a+{{x}^{2}} \right|\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{1}{2}\ln \left| \dfrac{a+1}{a} \right|=1\Leftrightarrow \left[ \begin{aligned}
& a=\dfrac{1}{{{e}^{2}}-1} \\
& a=-\dfrac{1}{{{e}^{2}}+1}\left( lọai \right) \\
\end{aligned} \right.$
Đáp án B.