Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ thỏa mãn...

-Đặt $si{{n}^{2}}x=t\Rightarrow dt=2\sin x.\cos x\text{d}x=2\cot x.si{{n}^{2}}x\text{d}x$
$\Rightarrow $ $\dfrac{\text{d}t}{t}=2\cot x\text{d}x$
-Đổi cận $\left\{ \begin{aligned}
& x=\dfrac{\pi }{4}\Rightarrow t=\dfrac{1}{2} \\
& x=\dfrac{\pi }{2}\Rightarrow t=1 \\
\end{aligned} \right.$
$\Rightarrow I=\dfrac{1}{2}\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{f(t)}{t}dt}=1$ $\Rightarrow $ $\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{f(t)}{t}dt}=2$ $\Rightarrow $ $ $ $\left( 1 \right)$
+)Xét $\int\limits_{1}^{16}{\dfrac{f\left( \sqrt{x} \right)}{x}dx}=1$
-Đặt $t=\sqrt{x}\Rightarrow {{t}^{2}}=x\Rightarrow 2tdt=dx$
-Đổi cận $\left\{ \begin{aligned}
& x=1\Rightarrow t=1 \\
& x=16\Rightarrow t=4 \\
\end{aligned} \right.$
Khi đó $\int\limits_{1}^{16}{\dfrac{f\left( \sqrt{x} \right)}{x}dx}=\int\limits_{1}^{4}{\dfrac{f(t)}{{{t}^{2}}}\text{d}{{t}^{2}}}=2\int\limits_{1}^{4}{\dfrac{f(t)}{t}\text{d}t}=1\Rightarrow \int\limits_{1}^{4}{\dfrac{f(t)}{t}\text{d}t}=\dfrac{1}{2}$
$\Rightarrow $ $$ $\left( 2 \right)$
Cộng theo vế 2 biểu thức $\left( 1 \right)$ và $\left( 2 \right)$ ta được $\int\limits_{\dfrac{1}{2}}^{4}{\dfrac{f(x)}{x}\text{d}}x=\dfrac{5}{2}$
+)Xét $k=\int\limits_{\dfrac{1}{8}}^{1}{\dfrac{f(4x)}{x}\text{d}x}=\int\limits_{\dfrac{1}{8}}^{1}{\dfrac{f(4x)}{4x}\text{d}\left( 4x \right)}$ đặt $t=4x$ $\left\{ \begin{aligned}
& x=1\Rightarrow t=4 \\
& x=\dfrac{1}{8}\Rightarrow t=\dfrac{1}{2} \\
\end{aligned} \right.$
$k=\int\limits_{\dfrac{1}{2}}^{4}{\dfrac{f(t)}{t}\text{d}t}=\int\limits_{\dfrac{1}{2}}^{4}{\dfrac{f(x)}{x}\text{d}x}=\dfrac{3}{2}$