Question

Calcule a integral

9) $\int\ {} \frac{dx}{x \sqrt{x^2+4} } \,$

Answer 1

Vamos utilizar a seguinte substituição trigonométrica:

x² + a² ⇔ x =  atg(u)

x² + 2² ⇔ x = 2tg(u)

Ou seja,

x = 2tg(u)

Vamos derivar ambos os lados...

dx = 2Sec²udu
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Logo,

$\\ = \int\limits { \frac{dx}{x \sqrt{x^2+4} } } \, = \int\limits { \frac{2Sec^2udu}{2tgu\sqrt{(2tgu)^2+4} } } \, \\ \\ = \int\limits { \frac{2Sec^2udu}{2tgu\sqrt{4tg^2u+4} } } \, \\ \\ = \int\limits { \frac{2Sec^2udu}{2tgu\sqrt{4(tg^2u+1)} } } \,$

Substituindo tg²u+1 = Sec²u
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$\\ = \int\limits { \frac{2sec^2udu}{2tgu \sqrt{4Sec^2u} } } \, \\ \\ = \int\limits { \frac{2sec^2udu}{2tgu*2Secu } } \, \\ \\ = \frac{1}{2} \int\limits { \frac{secudu}{tgu } } \, \\ \\ = \frac{1}{2} \int\limits { \frac{ \frac{1}{Cosu} du}{ \frac{Senu}{Cosu} } } \, \\ \\ = \frac{1}{2} \int\limits { \frac{1}{Cosu} } * \frac{Cosu}{Senu} \, du \\ \\ = \frac{1}{2} \int\limits { \frac{1}{Senu} } \, du$

Sabemos que, 1/Senu = Cossecu
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$= \frac{1}{2} \int\limits {Cossecu} \, du$

Sabemos que essa integral é tabelada e vale : ln| Cossecu - Cotgu|

Então teremos:

$\\ = \frac{1}{2} ln| Cossecu- Cotgu| + C$
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Achando o valor de cotgu e cossecu
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Temos que:

2tgu = x

$tgu = \frac{x}{2}$

Como tgu =  Co/ ca

logo,

Co = x
Ca = 2
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Achando hip:

hip² = co²+ca²

hip² = x² + 2²

hip = √(x²+4)
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Como Cotgu = 1/tgu

Então,

Cotgu = 1/(x/2)

Cotgu = 2/x
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Cossecu = 1/Senu

Cossecu = 1/(Co/hip)

Cossecu = Hip/Co

Cossecu = √(x²+4)/x
-------------------------

Agora só substituir:


$\\ = \frac{1}{2}ln| \frac{ \sqrt{x^2+4} }{x} - \frac{2}{x} |+C \\ \\ = \frac{1}{2}ln| \frac{ \sqrt{x^2+4}-2 }{x} |+C$

Answer 2

$\sf \displaystyle \int \frac{dx}{x\sqrt{x^2+4}}\\\\\\\int \frac{sec \left(u\right)}{2tan \left(u\right)}du\\\\\\=\frac{1}{2}\cdot \int \frac{sec \left(u\right)}{tan \left(u\right)}du\\\\\\=\frac{1}{2}\cdot \int \frac{1}{v}dv\\\\\\\frac{1}{2}ln \left|v\right|\\\\\\=\frac{1}{2}ln \left|tan \left(\frac{arctan \left(\frac{1}{2}x\right)}{2}\right)\right|\\\\\\\to \boxed{\sf =\frac{1}{2}ln \left|tan \left(\frac{arctan \left(\frac{1}{2}x\right)}{2}\right)\right|+C}$