What is $\int \frac{2}{4 + x^{2}} d x$ $int 2/(4+x^(2)) dx$ ?

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What is $\int \frac{2}{4 + x^{2}} d x$ $int 2/(4+x^(2)) dx$ ?

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Answer 1 · 2018-04-11T06:15:22

The answer is $= arctan (\frac{x}{2}) + C$ $=arctan(x/2)+C$

Explanation:

Perform this integral by substitution

$4 + x^{2} = 4 (1 + {(\frac{x}{2})}^{2})$ $4+x^2=4(1+(x/2)^2)$

Let $tan u = \frac{x}{2}$ $tanu=x/2$

${sec}^{2} u d u = \frac{1}{2} d x$ $sec^2udu=1/2dx$

$1 + {tan}^{2} u = {sec}^{2} u$ $1+tan^2u=sec^2u$

Therefore, the integral is

$\int \frac{2 d x}{4 + x^{2}} = \int \frac{4 {sec}^{2} u d u}{4 (1 + {tan}^{2} u)}$ $int(2dx)/(4+x^2)=int(4sec^2udu)/(4(1+tan^2u))$

$= \int \frac{{sec}^{2} u d u}{{sec}^{2} u}$ $=int(sec^2udu)/(sec^2u)$

$= \int (d u)$ $=int(du)$

$= u$ $=u$

$= arctan (\frac{x}{2}) + C$ $=arctan(x/2)+C$

Answer 2 · 2018-04-11T06:19:51

The integral is equal to $arctan (\frac{x}{2}) + C$ $arctan(x/2)+C$ .

Explanation:

To solve the integral, use the substitution $u = \frac{x}{2}$ $u=x/2$ , which means $x = 2 u$ $x=2u$ and $d x = 2 d u$ $dx=2du$ :

$= \int \frac{2}{4 + x^{2}}$ $color(white)=int2/(4+x^2)$ $d x$ $dx$

$= \int \frac{2}{4 + {(2 u)}^{2}}$ $=int2/(4+(2u)^2)$ $2 d u$ $2du$

$= \int \frac{4}{4 + 4 u^{2}}$ $=int4/(4+4u^2)$ $d u$ $du$

$=intcolor(red)cancelcolor(black)4/(color(red)cancelcolor(black)4(1+u^2))$ $du$

$=int1/(1+u^2)$ $du$

$=arctan(u)+C$

$=arctan(x/2)+C$

That's the integral. Hope this helped!

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What is $\int \frac{2}{4 + x^{2}} d x$ $int 2/(4+x^(2)) dx$ ?

2 Answers

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