How do you integrate $\int{\dfrac{dx}{{{x}^{2}}+25}}$ using substitution?

Hint: We can solve problems of integration like this recalling the Pythagorean identities in trigonometry. First, we start solving the problem assuming the variable $x$ to be $5\tan \theta $ and rewrite the given expression substituting the assumed value of $x$ . Then, we perform integration of the expression which will give us the integrated result in terms of $\theta $ . Again, we substitute the value of $\theta $ in the expression to get the result in terms of $x$ .

Complete step by step answer:
The given expression is
$\int{\dfrac{dx}{{{x}^{2}}+25}}$
To solve the integration by substituting trigonometric variables we must recall the Pythagorean identities in trigonometry.
\[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] could be useful if we see $a-b{{u}^{2}}$ (We try to get \[k(1-{{\sin }^{2}}\theta )=k{{\cos }^{2}}\theta \] )
In this case, we have a square added to a number.
Hence, we think about using the relation between $\tan \theta $ and $\sec \theta $ as
$1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ which gives us our substitution.
We want to have $25{{\tan }^{2}}\theta +25$ , so we will use $x=5\tan \theta $
Substituting the value of $x$ the denominator of the given expression becomes
$={{\left( 5\tan \theta \right)}^{2}}+25$
$=25{{\tan }^{2}}\theta +25$
Taking $25$ out common we get
$=25\left( {{\tan }^{2}}\theta +1 \right)$
$=25{{\sec }^{2}}\theta $
Also, to substitute the term $dx$ we differentiate $x=5\tan \theta $ as shown below
$\Rightarrow dx=d\left( 5\tan \theta \right)$
$\Rightarrow dx=5{{\sec }^{2}}\theta \cdot d\theta $
Also, from $x=5\tan \theta $ we get
$\theta ={{\tan }^{-1}}\left( \dfrac{x}{5} \right)$
Therefore, the integral becomes
$=\int{\dfrac{5{{\sec }^{2}}\theta d\theta }{25\left( {{\tan }^{2}}\theta +1 \right)}}$
$=\int{\dfrac{5{{\sec }^{2}}\theta d\theta }{25\left( {{\sec }^{2}}\theta \right)}}$
Taking out the constant terms we get
$=\dfrac{5}{25}\int{d\theta }$
$=\dfrac{1}{5}\int{d\theta }$
Doing the integration, we get
$=\dfrac{1}{5}\theta +c$
Here, $c$ is a constant.
Now, we substitute the value of $\theta $ in the above expression and get
$=\dfrac{1}{5}{{\tan }^{-1}}\left( \dfrac{x}{5} \right)+c$
Therefore, by integrating the given expression using trig substitution we get $\dfrac{1}{5}{{\tan }^{-1}}\left( \dfrac{x}{5} \right)+c$ .

Note:
While doing substitution using trigonometric relation, we must choose a proper trigonometric relation for substitution. Otherwise, we will not be getting the correct result. Also, while integrating using substitution we must also substitute the $dx$ term as it often gets ignored by students making the problem unsolvable.