Chọn phương án A.

Đặt \(u=\tan x\) ta có

• \(\mathrm{d}u=\left(1+\tan^2x\right)\mathrm{\,d}x=\left(1+u^2\right)\mathrm{\,d}x\)
  \(\Rightarrow\mathrm{\,d}x=\dfrac{1}{1+u^2}\mathrm{\,d}u\).
• \(x=0\Rightarrow u=\tan0=0\)
• \(x=\dfrac{\pi}{4}\Rightarrow u=\tan\dfrac{\pi}{4}=1\)

Khi đó $$\begin{eqnarray*}
&\displaystyle\int\limits_{0}^{\tfrac{\pi}{4}}f\left(\tan x\right)\mathrm{\,d}x&=4\\
\Leftrightarrow&\displaystyle\int\limits_{0}^{1}\dfrac{f(u)}{1+u^2}\mathrm{\,d}u&=4\\
\text{hay }&\displaystyle\int\limits_{0}^{1}\dfrac{f(x)}{1+x^2}\mathrm{\,d}x&=4.
\end{eqnarray*}$$
Ta lại có $$\begin{eqnarray*}
&\displaystyle\int\limits_{0}^{1}\dfrac{x^2\cdot f(x)}{x^2+1}\mathrm{\,d}x&=2\\
\Leftrightarrow&\displaystyle\int\limits_{0}^{1}\dfrac{\left(x^2+1\right)-1}{x^2+1}\cdot f(x)\mathrm{\,d}x&=2\\
\Leftrightarrow&\displaystyle\int\limits_{0}^{1}\left(1-\dfrac{1}{x^2+1}\right)\cdot f(x)\mathrm{\,d}x&=2\\
\Leftrightarrow&\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x-\displaystyle\int\limits_{0}^{1}\dfrac{f(x)}{x^2+1}\mathrm{\,d}x&=2\\
\Leftrightarrow&\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x-4&=2\\
\Leftrightarrow&\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x&=6.
\end{eqnarray*}$$