Question

int\frac{dx}{x^{2}-16

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Answer 1

Hope it will help you. ..

Answer 2

Topic :-

Integration

To Solve :-

$\displaystyle \int \dfrac{dx}{x^2-16}$

Formula to be Used :-

$\displaystyle \int \dfrac{dx}{x^2-a^2}=\dfrac{1}{2a}ln\mid{\dfrac{x-a}{x+a}}\mid+C$

Solution :-

$\displaystyle \int \dfrac{dx}{x^2-16}$

$\displaystyle \int \dfrac{dx}{x^2-(4)^2}$

Now, apply the formula,

here a = 4,

$\displaystyle \int \dfrac{dx}{x^2-4^2}=\dfrac{1}{2(4)}ln\mid{\dfrac{x-4}{x+4}}\mid+C$

$\displaystyle \int \dfrac{dx}{x^2-16}=\dfrac{1}{8}ln\mid{\dfrac{x-4}{x+4}}\mid+C$

Here, C is Constant of Integration

Additional Formulae :-

$\displaystyle \int \dfrac{dx}{a^2-x^2}=\dfrac{1}{2a}ln\mid{\dfrac{a+x}{a-x}}\mid+C$

$\displaystyle \int \dfrac{dx}{\sqrt{a^2-x^2}}=sin^{-1}{\dfrac{x}{a}}+C$

$\displaystyle \int \dfrac{dx}{{a^2+x^2}}=\dfrac{1}{a}tan^{-1}{\dfrac{x}{a}}+C$

$\displaystyle \int sinxdx=-cosx+C$

$\displaystyle \int cosxdx=sinx+C$