Integrate $\frac{1}{4 + x^{2}} d x$ $1/(4+x^2) dx$ ?

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Integrate $\frac{1}{4 + x^{2}} d x$ $1/(4+x^2) dx$ ?

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Answer 1 · 2018-05-31T13:08:42

$\int \frac{1}{4 + x^{2}} d x = \frac{1}{2} arctan (\frac{1}{2} x) + c$ $int1/(4+x^2)"d"x=1/2arctan(1/2x)+"c"$

Explanation:

$\int \frac{1}{4 + x^{2}} d x = \int \frac{1}{4 (1 + \frac{x^{2}}{4})} d x = \frac{1}{4} \int \frac{1}{1 + {(\frac{x}{2})}^{2}} d x$ $int1/(4+x^2)"d"x=int1/(4(1+x^2/4))"d"x=1/4int1/(1+(x/2)^2)"d"x$

Let $u = \frac{x}{2}$ $u=x/2$ and $d u = \frac{1}{2} d x$ $"d"u=1/2"d"x$

Then

$\frac{1}{4} \int \frac{1}{1 + {(\frac{x}{2})}^{2}} d x = \frac{1}{2} \int \frac{1}{1 + u^{2}} d u$ $1/4int1/(1+(x/2)^2)"d"x=1/2int1/(1+u^2)"d"u$

This is a well-known integral which evaluates to

$\frac{1}{2} arctan u = \frac{1}{2} arctan (\frac{1}{2} x) + c$ $1/2arctanu=1/2arctan(1/2x)+"c"$

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Integrate $\frac{1}{4 + x^{2}} d x$ $1/(4+x^2) dx$ ?

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