Chọn A

Ta có $I=\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{3}}{\int }}\frac{f\left(\tan x\right)}{co{s}^{2}x}d\text{x}+\underset{0}{\overset{\sqrt{e-1}}{\int }}\frac{xf\left(\ln \left(x^2+1\right)\right)}{x^2+1}d\text{x=J+K}$.

+)$J=\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{3}}{\int }}\frac{f\left(\tan x\right)}{co{s}^{2}x}d\text{x}$. Đặt $t=\tan x\Rightarrow dt=\frac{1}{co{s}^{2}x}dx$. Đổi cận $x=\frac{\pi }{3}\Rightarrow t=\sqrt{3};x=\frac{\pi }{4}\Rightarrow t=1$.

Suy ra $J=\underset{1}{\overset{\sqrt{3}}{\int }}f\left(t\right)dt=\underset{1}{\overset{\sqrt{3}}{\int }}f\left(x\right)dx=\underset{1}{\overset{\sqrt{3}}{\int }}\left(2x^3-x\right)dx={\left.\left(\frac{x^4}{2}-\frac{x^2}{2}\right)\right|}_1^{\sqrt{3}}=3$.

+) $K=\underset{0}{\overset{\sqrt{e-1}}{\int }}\frac{xf\left(\ln \left(x^2+1\right)\right)}{x^2+1}dx$. Đặt $t=\ln \left(x^2+1\right)\Rightarrow dt=\frac{2x}{x^2+1}dx\Rightarrow \frac{x}{x^2+1}dx=\frac{dt}{2}$

Đổi cận $x=\sqrt{e-1}\Rightarrow t=1;x=0\Rightarrow t=0$.

Suy ra $K=\underset{0}{\overset{1}{\int }}f\left(t\right)\frac{dt}{2}=\underset{0}{\overset{1}{\int }}f\left(x\right)\frac{dx}{2}=\underset{0}{\overset{1}{\int }}\frac{-3x+4}{2}dx={\left.\left(-\frac{3}{4}{x}^2+2x\right)\right|}_0^1=\frac{5}{4}$