How do you evaluate the integral $\int {x{{\sec }^2}xdx} $?

Hint: To solve this integral, you need to know the integration by parts method and the formula for that is $\int {udv = uv - \int {vdu} } $ and we should now the basic formula such as$\int {{{\sec }^2}xdx = \tan x} $ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$.

Complete step by step answer:
Let us consider the given solution:
$\int {x{{\sec }^2}xdx} $
Let’s consider $I = \int {x{{\sec }^2}xdx} $
Let us apply $\int {udv} $ formula to solve this equation, for that let us consider $u = x$ and $dv = {\sec ^2}xdx$ and the formula for integration by parts is$\int {udv = uv - \int {vdu} } $, if you look at the formula it contains $du$ and $v$, to find it we have to differentiate $u$ and integrate $dv$ we get,
$u = x,du = dx$ and $dv = {\sec ^2}xdx$, $v = \tan x$, substituting the value in the formula we get,
$
\Rightarrow I = \int {x{{\sec }^2}x = x\tan x - \int {\tan x dx} } \\
\Rightarrow I = x\tan x - \int {\dfrac{{\sin x}}{{\cos x}}dx} \\
\Rightarrow I = x\tan x + \int {\dfrac{{ - \sin x}}{{\cos x}}dx} \\
 $
Let $t = \cos x$ and the by differentiating $t$ we get,
$ \Rightarrow dt = - \sin x dx$, by substituting the values, we get,
$ \Rightarrow I = x\tan x + \int {\dfrac{1}{t}dt} $
$ \Rightarrow I = x\tan x + \int {\dfrac{1}{t}dt} $
$ \Rightarrow I = x\tan x + \log t + C$
This is the required solution.

Note: In $I = x\tan x + \log t + C$, where C is the constant, whenever we integrate a logarithm, we will get a constant to balance the LHS and RHS. The formula for $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and the $\cot x$ will be the reciprocal of $\tan x$, which will be equal to $\cot x = \dfrac{{\cos x}}{{\sin x}}$. And know the other identities of the trigonometry also. Because in many of the materials, they don’t solve every step, they will just skip some of the steps and proceed to the answer. So knowing the basic identities will help you to follow all the steps in your materials.
This is an indefinite integral, while solving a indefinite integral we should be more careful. Because this is the most complicated and important problem which will be continuously used in your higher studies. Never hesitate to solve more problems. Solving more problems will relieve you from difficulties.