\(I_{6} =\int x{\rm tan}^{2} x{\rm d}x =\int x\frac{\sin ^{2} x}{{\rm cos}^{2} x} {\rm d}x= \int x\frac{1-{\rm cos}^{2} x}{{\rm cos}^{2} x} {\rm d}x= \int x.\frac{1}{{\rm cos}^{2} x} {\rm d}x -\int x{\rm d}x \)
Đặt \(\left\{\begin{array}{l} {u=x} \\ {{\rm d}v=\frac{1}{{\rm cos}^{2} x} {\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u={\rm d}x} \\ {v=\tan x} \end{array}\right. \)
Suy ra
\(I_{6} =x\tan x-\int \tan x{\rm d}x -\int x{\rm d}x =x\tan x-\int \frac{\sin x}{{\rm cos}x} {\rm d}x -\int x{\rm d}x \)
\(=x\tan x+\int \frac{d\left(c{\rm os}x\right)}{{\rm cos}x} -\int x{\rm d}x =x\tan x+\ln \left|{\rm cos}x\right|-\frac{x^{2} }{2} +C.\)