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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& {{x}^{2}}+x-1\text{ khi }x\ge 0 \\
& 2{{x}^{2}}-1\text{ khi }x<0 \\
\end{aligned} \right. $. Tính $ \int\limits_{\dfrac{1}{e}}^{e}{\dfrac{{f}'\left( \ln x \right)\ln x}{x}}dx$.
A. $\dfrac{5}{2}\cdot $
B. $\dfrac{7}{2}\cdot $
C. $\dfrac{3}{2}\cdot $
D. $\dfrac{1}{2}\cdot $
& {{x}^{2}}+x-1\text{ khi }x\ge 0 \\
& 2{{x}^{2}}-1\text{ khi }x<0 \\
\end{aligned} \right. $. Tính $ \int\limits_{\dfrac{1}{e}}^{e}{\dfrac{{f}'\left( \ln x \right)\ln x}{x}}dx$.
A. $\dfrac{5}{2}\cdot $
B. $\dfrac{7}{2}\cdot $
C. $\dfrac{3}{2}\cdot $
D. $\dfrac{1}{2}\cdot $
Đặt $t=\ln x\Rightarrow dt=\dfrac{1}{x}dx$
$\Rightarrow I=\int\limits_{\dfrac{1}{e}}^{e}{\dfrac{{f}'\left( \ln x \right)\ln x}{x}}dx=\int\limits_{-1}^{1}{t.{f}'\left( t \right)}dt=\int\limits_{-1}^{1}{t}d\left( f\left( t \right) \right)=\left. t.f\left( t \right) \right|_{-1}^{1}-\int\limits_{-1}^{1}{f\left( t \right)}dt=f\left( 1 \right)+f\left( -1 \right)+\int\limits_{-1}^{1}{f\left( t \right)}dt$ Ta có $f\left( 1 \right)={{1}^{2}}+1-1=1$ ; $f\left( -1 \right)=2.{{\left( -1 \right)}^{2}}-1=1$ ;
$\int\limits_{-1}^{1}{f\left( t \right)}dt=\int\limits_{-1}^{1}{f\left( x \right)}dx=\int\limits_{-1}^{0}{f\left( x \right)}dx+\int\limits_{0}^{1}{f\left( x \right)}dx=\int\limits_{-1}^{0}{\left( 2{{x}^{2}}-1 \right)}dx+\int\limits_{0}^{1}{\left( {{x}^{2}}+x-1 \right)}dx=-\dfrac{1}{2}$
$\Rightarrow I=f\left( 1 \right)+f\left( -1 \right)+\int\limits_{-1}^{1}{f\left( t \right)}dt=1+1-\dfrac{1}{2}=\dfrac{3}{2}$
Đổi cận
| $x$ | $\dfrac{1}{e}$ | $e$ |
| $t$ | $-1$ | $1$ |
$\int\limits_{-1}^{1}{f\left( t \right)}dt=\int\limits_{-1}^{1}{f\left( x \right)}dx=\int\limits_{-1}^{0}{f\left( x \right)}dx+\int\limits_{0}^{1}{f\left( x \right)}dx=\int\limits_{-1}^{0}{\left( 2{{x}^{2}}-1 \right)}dx+\int\limits_{0}^{1}{\left( {{x}^{2}}+x-1 \right)}dx=-\dfrac{1}{2}$
$\Rightarrow I=f\left( 1 \right)+f\left( -1 \right)+\int\limits_{-1}^{1}{f\left( t \right)}dt=1+1-\dfrac{1}{2}=\dfrac{3}{2}$
Đáp án C.