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Evaluate the given integral $\int {\dfrac{1}{{{x^2} + 16}}dx} $

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Hint: - Here we put into formula to solve easily$\left( {\int {\dfrac{1}{{{a^2} + {x^2}}}dx = \dfrac{1}{a}{{\tan }^{ - 1}}\dfrac{x}{a} + c} } \right)$.

$\int {\dfrac{1}{{{x^2} + 16}}dx} {\text{ }}$
We will use a substitution method to solve this problem, you may put a direct formula and get an answer.
Put $x{\text{ = 4tan}}\theta $
Differentiate with respect to$x$.
$dx = 4{\sec ^2}\theta d\theta $
Now,
$\int {\dfrac{{dx}}{{16 + {x^2}}}} = \int {\dfrac{{4{{\sec }^2}\theta d\theta }}{{16 + 16{{\tan }^2}\theta }} = \int {\dfrac{{4{{\sec }^2}\theta d\theta }}{{16\left( {1 + {{\tan }^2}\theta } \right)}}} } $
$\left( {\because 1 + {{\tan }^2}\theta = {{\sec }^2}\theta } \right)$
So, after cancel out, we get
$ \Rightarrow \dfrac{1}{4}\int {d\theta = \dfrac{1}{4}\theta + c = } \dfrac{1}{4}{\tan ^{ - 1}}\dfrac{x}{4} + c$ $\left( \begin{gathered}
  \because x = 4\tan \theta \\
  \therefore \theta = {\tan ^{ - 1}}\dfrac{x}{4} \\
\end{gathered} \right)$
Therefore the required answer is $\dfrac{1}{4}{\tan ^{ - 1}}\dfrac{x}{4} + c$
Note: - Whenever we face such type of integral question, we have to use substitution method to solve easily or this question can be solve by using direct formula$\left( {\int {\dfrac{1}{{{a^2} + {x^2}}}dx = \dfrac{1}{a}{{\tan }^{ - 1}}\dfrac{x}{a} + c} } \right)$.
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