Mathematics

# Evaluate the given integral.$\displaystyle\int { \cfrac { 1 }{ { x }^{ 2 }+16 } } dx\quad$

##### SOLUTION
$I=\displaystyle\int{\dfrac{dx}{{x}^{2}+16}}$

$=\displaystyle\int{\dfrac{dx}{{x}^{2}+{4}^{2}}}$

We know that $\displaystyle\int{\dfrac{dx}{{x}^{2}+{a}^{2}}}=\dfrac{1}{a}{\tan}^{-1}{\dfrac{x}{a}}+c$ ..........where $c$ is constant of integration

$=\dfrac{1}{4}{\tan}^{-1}{\dfrac{x}{4}}+c$     ..........where $c$ is constant of integration

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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