Cho hàm số f(x) liên tục trên \( \left[ \frac{2}{5};1 \right] \) và thỏa mãn \( 2f(x)+5f\left( \frac{2}{5x} \right)=3x,\text{ }\forall x\in \left[ \frac{2}{5};1 \right] \). Khi đó \( I=\int\limits_{\frac{2}{15}}^{\frac{1}{3}}{\ln 3x.{f}'(3x)dx} \) bằng

Hướng dẫn giải:

Đáp án B.

Ta có:  \( 2f(x)+5f\left( \frac{2}{5x} \right)=3x,\text{ }\forall x\in \left[ \frac{2}{5};1 \right] \)

 \( \Leftrightarrow 2\frac{f(x)}{x}+5\frac{f\left( \frac{2}{5x} \right)}{x}=3\Rightarrow 2\int\limits_{\frac{2}{5}}^{1}{\frac{f(x)}{x}dx}+5\int\limits_{\frac{2}{5}}^{1}{\frac{f\left( \frac{2}{5x} \right)}{x}dx}=\int\limits_{\frac{2}{5}}^{1}{3dx}=\frac{9}{5} \)  (*).

+ Xét \({{I}_{1}}=5\int\limits_{\frac{2}{5}}^{1}{\frac{f\left( \frac{2}{5x} \right)}{x}dx}\).

Đặt  \( t=\frac{2}{5x}\Rightarrow \left\{ \begin{align}  & x=\frac{2}{5t} \\ & {{t}^{2}}=\frac{4}{25{{x}^{2}}}\Rightarrow \frac{25}{4}{{t}^{2}}=\frac{1}{{{x}^{2}}} \\  & dt=-\frac{2}{5{{x}^{2}}}dx=-\frac{2}{5}.\frac{25}{4}{{t}^{2}}dx\Rightarrow -\frac{2}{5}\frac{dt}{{{t}^{2}}}=dx \\ \end{align} \right. \).

Đổi cận:  \( \left\{ \begin{align}  & x=\frac{2}{5}\Rightarrow t=1 \\  & x=1\Rightarrow t=\frac{2}{5} \\ \end{align} \right. \).

Khi đó: \({{I}_{1}}=5\int\limits_{\frac{2}{5}}^{1}{\frac{f\left( \frac{2}{5x} \right)}{x}dx}=5\int\limits_{1}^{\frac{2}{5}}{\frac{f(t)}{\frac{2}{5t}}.\left( -\frac{2}{5}\frac{dt}{{{t}^{2}}} \right)}=5\int\limits_{\frac{2}{5}}^{1}{\frac{f(t)}{t}dt}=5\int\limits_{\frac{2}{5}}^{1}{\frac{f(x)}{x}dx}\).

Từ (*) suy ra:  \( 2\int\limits_{\frac{2}{5}}^{1}{\frac{f(x)}{x}dx}+5\int\limits_{\frac{2}{5}}^{1}{\frac{f(x)}{x}dx}=\frac{9}{5}\Leftrightarrow \int\limits_{\frac{2}{5}}^{1}{\frac{f(x)}{x}dx}=\frac{9}{35} \).

+ Xét  \( I=\int\limits_{\frac{2}{15}}^{\frac{1}{3}}{\ln 3x.{f}'(3x)dx} \).

Đặt  \( t=3x\Rightarrow dt=3dx\Rightarrow dx=\frac{1}{3}dt  \).

Đổi cận:  \( \left\{ \begin{align} & x=\frac{2}{15}\Rightarrow t=\frac{2}{5} \\  & x=\frac{1}{3}\Rightarrow t=1 \\ \end{align} \right. \).

 \( \Rightarrow I=\frac{1}{3}\int\limits_{\frac{2}{5}}^{1}{\ln t.{f}'(t)dt} \).

Đặt:  \( \left\{ \begin{align}  & u=\ln t\Rightarrow du=\frac{1}{t}dt \\  & dv={f}'(t)dt\Rightarrow v=f(t) \\ \end{align} \right. \).

 \( I=\left. \frac{1}{3}\ln t.f(t) \right|_{\frac{2}{5}}^{1}-\frac{1}{3}\int\limits_{\frac{2}{5}}^{1}{\frac{f(t)}{t}dt}=-\frac{1}{3}\ln \frac{2}{5}.f\left( \frac{2}{5} \right)-\frac{3}{35} \).

+ Xét  \( 2f(x)+5f\left( \frac{2}{5x} \right)=3x,\text{ }\forall x\in \left[ \frac{2}{5};1 \right] \).

Thay  \( x=1;\text{ }x=\frac{2}{5} \) vào biểu thức trên, ta được hệ phương trình sau:

 \( \left\{ \begin{align}  & 2f(1)+5f\left( \frac{2}{5} \right)=3 \\  & 2f\left( \frac{2}{5} \right)+5f(1)=\frac{6}{5} \\ \end{align} \right.\) \( \Leftrightarrow \left\{ \begin{align}  & f(1)=0 \\  & f\left( \frac{2}{5} \right)=\frac{3}{5} \\ \end{align} \right. \) .

Suy ra:  \( I=-\frac{1}{3}\ln \frac{2}{5}.\frac{3}{5}-\frac{3}{35}=\frac{1}{5}\ln \frac{5}{2}-\frac{3}{35} \).