How do you integrate $ \int \dfrac{x}{{\sqrt {{x^2} - 7} }}dx $ by trigonometric substitution?

Hint: To solve this question, first we will assume any of the set of variables or constant be another variable to get the expression easier. And conclude until the non-operational state is not achieved. And finally substitute the assumed value. It is also called integration by substitution.

Complete step by step solution:
The given expression: $ \int \dfrac{x}{{\sqrt {{x^2} - 7} }}dx $
We can integrate this expression by the substitution-
Let $ {x^2} - 7 = t $ .
Now, differentiate the above assumed equation:
 $ \Rightarrow \dfrac{{d{x^2}}}{{dx}} - \dfrac{d}{{dx}}(7) = \dfrac{{dt}}{{dx}} $
 $ \Rightarrow 2x - 0 = \dfrac{{dt}}{{dx}} $
 $ \Rightarrow 2x.dx = dt $
 $ \Rightarrow x.dx = \dfrac{{dt}}{2} $
Now, use the above equation in the main expression:
 $
  \because \int \dfrac{x}{{\sqrt {{x^2} - 7} }}dx \;
  $
put $ \dfrac{{dt}}{2} $ instead of $ x.dx $ .
 $
   = \int \dfrac{{dt}}{2}\dfrac{1}{{\sqrt t }} \\
   = \dfrac{1}{2}\int {t^{ - \dfrac{1}{2}}}dt \\
   = \dfrac{1}{2}\dfrac{{{t^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + C \\
   = \sqrt t + C \;
  $
Now, substitute the actual value of $ t $ :
 $ = \sqrt {{x^2} - 7} + C $
Hence, the integration of
 $
  \int \dfrac{x}{{\sqrt {{x^2} - 7} }}dx \;
  $ is $ \sqrt {{x^2} - 7} + C $ .
So, the correct answer is “ $ \sqrt {{x^2} - 7} + C $ ”.

Note: Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.