Evaluate $\int {\dfrac{{{e^x}(1 + x)}}{{{{\cos }^2}(x{e^x})}}dx = } $

Hint:In this question first let us suppose that the $x{e^x} = z$, on differentiating with respect to x we get ${e^x}(x + 1)dx = dz$ . Now put it in this equation, the integration become $\int {{{\sec }^2}zdz} $ now integrate it and at last put the value of $z$.

Complete step-by-step answer:
In the given question we have to find the value of $\int {\dfrac{{{e^x}(1 + x)}}{{{{\cos }^2}(x{e^x})}}dx} $
Hence for this let us suppose that the $x{e^x} = z$
Now differentiate this on both side with respect to $x$
As we know that the if the function is in multiplication then differentiation of this is $u.v = u'v + u.v'$
$x\dfrac{{d{e^x}}}{{dx}} + {e^x}\dfrac{{dx}}{{dx}} = \dfrac{{dz}}{{dx}}$
As in this $\dfrac{{d{e^x}}}{{dx}} = {e^x}$ and $\dfrac{{dx}}{{dx}} = 1$
So the remaining $x.{e^x} + {e^x} = \dfrac{{dz}}{{dx}}$
or $(x.{e^x} + {e^x})dx = dz$, ${e^x}(x + 1)dx = dz$
put this value in the equation $\int {\dfrac{{{e^x}(1 + x)}}{{{{\cos }^2}(x{e^x})}}dx} $ we get
$\int {\dfrac{{dz}}{{{{\cos }^2}(z)}}} $
$\int {{{\sec }^2}zdz} $
Hence the integration become $\int {{{\sec }^2}zdz} $
Integration of ${\sec ^2}z$ is $\tan z$
so $\int {{{\sec }^2}zdz} $ = $\tan z + C$ where C is constant
Now put the value of $x{e^x} = z$that is $\tan x{e^x} + C$
therefore $\int {\dfrac{{{e^x}(1 + x)}}{{{{\cos }^2}(x{e^x})}}dx} $ = $\tan x{e^x} + C$

Note:Some of the integral that we have to remember for solving these types of questions are
$\
  \int {\tan xdx = \ln \left| {\sec x} \right| + C} \\
  \int {\cot xdx = \ln \left| {\sin x} \right| + C} \\
  \int {\sec xdx = \ln \left| {\sec x + \tan x} \right| + C} \\
  \int {\cos ecxdx = \ln \left| {\cos ecx - \cot x} \right| + C} \\
  \int {\sec x.\tan xdx = \sec x + C} \\
  \int {{{\tan }^2}x} dx = \tan x - x + C \\
\ $
 Definite Integral represents the area under that curve according to that limit .