Tính \[I = \smallint x{\tan ^2}xdx\] ta được:

\[I = \smallint x{\tan ^2}xdx = \smallint x\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)dx = \smallint x.\frac{1}{{{{\cos }^2}x}}dx - \smallint xdx = {I_1} - {I_2}\]

Ta có:\[{I_2} = \smallint xdx = \frac{{{x^2}}}{2} + {C_2},{I_1} = \smallint x\frac{1}{{{{\cos }^2}x}}dx\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = \frac{1}{{co{s^2}x}}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = tanx}\end{array}} \right.\)

\[\begin{array}{*{20}{l}}{ \Rightarrow {I_1} = x\tan x - \smallint \tan xdx + {C_1} = x\tan x - \smallint \frac{{\sin x}}{{\cos x}}dx + {C_1}}\\{ = x\tan x + \smallint \frac{{d\left( {\cos x} \right)}}{{\cos x}} + {C_1} = x\tan x + \ln \left| {\cos x} \right| + {C_1}.}\\{ \Rightarrow I = x\tan x + \ln \left| {\cos x} \right| + {C_1} - \frac{{{x^2}}}{2} - {C_2} = x\tan x + \ln \left| {\cos x} \right| - \frac{{{x^2}}}{2} + C.}\end{array}\]

Đáp án cần chọn là: A