Đáp án:
Giải thích các bước giải:
Đặt:
\(\begin{array}{l}
\begin{array}{*{20}{c}}
u& = &x& \Rightarrow &{du}& = &{dx}\\
{dv}& = &{({{\tan }^2}x + 1 - 1)dx}& \Rightarrow &v& = &{\tan x - x}
\end{array}\\
I = x(\tan x - x) - \smallint \tan x.dx + \smallint x.dx = x.\tan x - \frac{{{x^2}}}{2} - J\\
J = \smallint \tan x.dx = \smallint \frac{{\sin x}}{{\cos x}}dx = - \smallint \frac{{d(\cos x)}}{{\cos x}} = - \ln |\cos x|\\
\to I = x.\tan x - \frac{{{x^2}}}{2} + \ln |\cos x| + C\\
\to \int\limits_0^{\frac{\pi }{4}} {x.{{\tan }^2}xdx = } x.\tan x\left| {_0^{\frac{\pi }{4}}} \right. - \frac{{{x^2}}}{2}\left| {_0^{\frac{\pi }{4}}} \right. + \ln |\cos x|\left| {_0^{\frac{\pi }{4}}} \right.\\
= \frac{\pi }{4} - \frac{{{\pi ^2}}}{{32}} + \ln \frac{{\sqrt 2 }}{2}
\end{array}\)
