Cho hàm số $\Large f(x)$ liên tục trên $\Large \mathbb{R}$ và thoả mãn

Cho hàm số $\Large f(x)$ liên tục trên $\Large \mathbb{R}$ và thoả mãn $\Large \int\limits_0^{\frac{\pi}{4}}tanx.f(cos^2x).\mathrm{d}x=1$, $\Large \int\limits_e^{e^2}\dfrac{f(\mathrm{ln}^2x)}{x\mathrm{ln}x}\mathrm{d}x=1$. Tính tích phân $\Large \int\limits_{\frac{1}{4}}^2\dfrac{f(2x)}{x}\mathrm{d}x$.

• A.

  $\Large I=1$.

• B.

  $\Large I=4$.

• C.

  $\Large I=3$.

• D.

  $\Large I=2$.

Chọn B
Xét $\Large A=\int\limits_0^{\frac{\pi}{4}}tanx.f(cos^2x).\mathrm{d}x=1$.

Đặt $\Large t=cos^2x$ $\Large \Rightarrow \mathrm{d}t=2cosx.(-sinx).\mathrm{d}x$ $\Large \Rightarrow \mathrm{d}t=-2cos^2x.tanx\mathrm{d}x$ $\Large tanx.\mathrm{d}x=-\dfrac{\mathrm{d}t}{2t}$.

Đổi cận $\Large x=0$ $\Large \Rightarrow t=1; x=\dfrac{\pi}{4}$ $\Large \Rightarrow t=\dfrac{1}{2}$.

Ta được $\Large A=\int\limits_0^{\frac{\pi}{4}}tanx.f(cos^2x).\mathrm{d}x$ $\Large =\int\limits_1^{\frac{1}{2}}f(t).\left(-\dfrac{\mathrm{d}t}{2t}\right)$ $\Large =\dfrac{1}{2}\int\limits_{\frac{1}{2}}^1\dfrac{f(t)}{t}.\mathrm{d}t$ $\Large =\dfrac{1}{2}\int\limits_{\frac{1}{2}}^1\dfrac{f(x)}{x}\mathrm{d}x$

$\Large \Rightarrow \int\limits_{\frac{1}{2}}^1\dfrac{f(x)}{x}.\mathrm{d}x=2A=2$.

Xét $\Large B=\int\limits_e^{e^2}\dfrac{f(\mathrm{ln}^2x)}{x\mathrm{ln}x}.\mathrm{d}x=1$

Đặt $\Large t=\mathrm{ln}^2x\Rightarrow \mathrm{d}t=2\mathrm{ln}x.\dfrac{1}{x}.\mathrm{d}x$ $\Large \Rightarrow \dfrac{1}{x\mathrm{ln}x}.\mathrm{d}x=\dfrac{\mathrm{d}t}{2t}$.

Đổi cận $\Large x=e$ $\Large \Rightarrow t=1; x=e^2$ $\Large \Rightarrow t=4$.

Ta được $\Large B=\int\limits_e^{e^2}\dfrac{f(\mathrm{ln}^2x)}{x\mathrm{ln}x}\mathrm{d}x$ $\Large =\int\limits_1^4f(t).\dfrac{\mathrm{d}t}{2t}$ $\Large =\dfrac{1}{2}\int\limits_1^4\dfrac{f(x)}{x}\mathrm{d}x$ $\Large \Rightarrow \int\limits_1^4\dfrac{f(x)}{x}.\mathrm{d}x=2B=2$.

Mặt khác, ta có $\Large \int\limits_{\frac{1}{4}}^2\dfrac{f(2x)}{x}.\mathrm{d}x$. Đặt $\Large t=2x\Rightarrow \mathrm{d}t=2\mathrm{d}x$ $\Large \Rightarrow \mathrm{d}x=\dfrac{\mathrm{d}t}{2}$ và $\Large x=\dfrac{t}{2}$.

Đổi cận $\Large x=\dfrac{1}{4}$ $\Large \Rightarrow t=\dfrac{1}{2}; x=2$ $\Large \Rightarrow t=4$.

Ta được $\Large \int\limits_{\frac{1}{4}}^2\dfrac{f(2x)}{x}\mathrm{d}x$ $\Large =\int\limits_{\frac{1}{2}}^4\dfrac{f(t)}{\dfrac{t}{2}}.\dfrac{\mathrm{d}t}{2}$ $\Large =\int\limits_{\frac{1}{2}}^4\dfrac{f(x)}{x}.\mathrm{d}x$ $\Large =\int\limits_{\frac{1}{2}}^1\dfrac{f(x)}{x}.\mathrm{d}x+\int\limits_1^4\dfrac{f(x)}{x}.\mathrm{d}x$ $\Large =2+2=4$.

Vậy $\Large \int\limits_{\frac{1}{4}}^2\dfrac{f(2x)}{x}.\mathrm{d}x=4$.