Evaluate the indefinite integral. ∫sec2(x)dxx

Step 1
We have to evaluate the integral:
${\displaystyle \int \frac{{\sec }^2\left(\sqrt{x}\right)dx}{\sqrt{x}}}$
We will use substitution method since derivatives of one function is present in the integral.
So assuming,
${\displaystyle t=\sqrt{x}}$
differentiating with respect to x,
${\displaystyle \frac{dt}{dx}=\frac{d\sqrt{x}}{dx}}$
${\displaystyle \frac{dt}{dx}=\frac{1}{2\sqrt{x}}}$
${\displaystyle 2dt=\frac{dx}{\sqrt{x}}}$
Substituting above values, we get
${\displaystyle \int \frac{{\sec }^2\left(\sqrt{x}\right)dx}{\sqrt{x}}=\int {\sec }^2\left(t\right)\left(2dt\right)}$
${\displaystyle =2\int {\sec }^2\left(t\right)dt}$
Step 2
Since we know that
${\displaystyle \int {\sec }^{2}xdx=\tan x+C}$
therefore,
${\displaystyle 2\int {\sec }^2\left(t\right)dt=2\tan \left(t\right)+C}$
${\displaystyle =2\tan \left(\sqrt{x}\right)+C}$
Where, C is an arbitrary constant.
Hence, value of given integral is ${\displaystyle 2\tan \left(\sqrt{x}\right)+C}$.

${\displaystyle \int \frac{{\sec }^2\left(\sqrt{x}\right)}{\sqrt{x}}dx}$
Substitution ${\displaystyle u=\sqrt{x}\Rightarrow \frac{du}{dx}=\frac{1}{2\sqrt{x}}\Rightarrow dx=2\sqrt{x}du}$
${\displaystyle =2\int {\sec }^2\left(u\right)du}$
${\displaystyle \int {\sec }^2\left(u\right)du}$
This is the well-known tabular integral:
${\displaystyle =\tan \left(u\right)}$
We substitute the already calculated integrals:
${\displaystyle 2\int {\sec }^2\left(u\right)du}$
${\displaystyle =2\tan \left(u\right)}$
Reverse replacement ${\displaystyle u=\sqrt{x}:}$
${\displaystyle =2\tan \left(\sqrt{x}\right)}$
Solution:
${\displaystyle =2\tan \left(\sqrt{x}\right)+C}$

$\int \frac{\sec (\sqrt{x}{)}^2}{\sqrt{x}}dx$
$\int 2\sec (t{)}^{2}dt$
$2\times \int \sec (t{)}^{2}dt$
$2\tan (t)$
$2\tan (\sqrt{x})$
Add $C\in \mathbb{R}$
Solution:
$2\tan (\sqrt{x})+C,C\in \mathbb{R}$