Hàm số \(y=f(x)\) liên tục trên \([1;4]\) và thỏa mãn \(f(x)=\dfrac{f\left(2\sqrt{x}-1\right)}{\sqrt{x}}+\dfrac{\ln x}{x}\). Tính tích phân \(I=\displaystyle\int\limits_{3}^{4}f(x)\mathrm{\,d}x\).
 | \(I=3+2\ln^22\) |
 | \(I=\ln^2\) |
 | \(I=2\ln2\) |
 | \(I=2\ln^22\) |
Chọn phương án D.
\displaystyle\int\limits_{1}^{4}f(x)\mathrm{\,d}x&=\displaystyle\int\limits_{1}^{4}\dfrac{f\left(2\sqrt{x}-1\right)}{\sqrt{x}}\mathrm{\,d}x+\displaystyle\int\limits_{1}^{4}\dfrac{\ln x}{x}\mathrm{\,d}x\\
&=M+N.
\end{aligned}$$
♥ Đặt \(u=2\sqrt{x}-1\) ta có
- \(\mathrm{\,d}u=\dfrac{1}{\sqrt{x}}\mathrm{\,d}x\)
- \(x=1\Rightarrow u=1\)
- \(x=4\Rightarrow u=3\)
Khi đó ta có $$\begin{aligned}
M=\displaystyle\int\limits_{1}^{4}\dfrac{f\left(2\sqrt{x}-1\right)}{\sqrt{x}}\mathrm{\,d}x&=\displaystyle\int\limits_{1}^{3}f(u)\mathrm{\,d}u\\
&=\displaystyle\int\limits_{1}^{3}f(x)\mathrm{\,d}x.\quad(1)
\end{aligned}$$
♥ Đặt \(v=\ln x\) ta có
- \(\mathrm{\,d}v=\dfrac{1}{x}\mathrm{\,d}x\)
- \(x=1\Rightarrow v=\ln1=0\)
- \(x=4\Rightarrow v=\ln4=2\ln2\)
Khi đó ta có $$\begin{aligned}
M=\displaystyle\int\limits_{1}^{4}\dfrac{\ln x}{x}\mathrm{\,d}x&=\displaystyle\int\limits_{0}^{2\ln2}v\mathrm{\,d}v\\
&=\dfrac{v^2}{2}\bigg|_0^{2\ln2}=2\ln^22.\quad(2)
\end{aligned}$$
Từ (1) và (2) ta có $$\begin{eqnarray*}
&\displaystyle\int\limits_{1}^{4}f(x)\mathrm{\,d}x&=\displaystyle\int\limits_{1}^{3}f(x)\mathrm{\,d}x+2\ln^22\\
\Leftrightarrow&\displaystyle\int\limits_{1}^{4}f(x)\mathrm{\,d}x-\displaystyle\int\limits_{1}^{3}f(x)\mathrm{\,d}x&=2\ln^22\\
\Leftrightarrow&\displaystyle\int\limits_{3}^{4}f(x)\mathrm{\,d}x&=2\ln^22.
\end{eqnarray*}$$