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
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