Question

Calcule a integral por substituição de variável

17) $\int\ {} \frac{dx}{ (\frac{1}{3}x-8)^5 } \,$

Answer 1

Olá lucas:

Façamos,

$u = \frac{1}{3} x-8$

Derivando implicitamente em "u"



$\\ \frac{d}{du} (u) = \frac{d}{du} ( \frac{1}{3}x-8) \\ \\ 1 = ( \frac{1}{3} +0) \frac{dx}{du} \\ \\ 1du = \frac{1}{3} dx \\ \\ 3du = dx$

Agora, substituindo-se:

$\left \{ {{u= \frac{1}{3}x-8 } \atop {dx=3du}} \right.$

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$\\ \int\limits \frac{dx}{( \frac{1}{3} x-8)^5} {} \, = \int\limits { \frac{3du}{u^5} } \, \\ \\ = 3 \int\limits { \frac{du}{u^5} \\ \\ = 3 \int\limits u^-^5du$

Observe, que n ≠ -1

Então podemos utilizar a integração por potência:

$n \neq -1, \left \{ {{ \int\limits {x^n} \, dx = \frac{x^n^+^1}{n+1} } \right.$

Logo,


$\\ 3\int\limits u^-^5{} \, du = \frac{3u^-^5^+^1}{-5+1} \\ \\ = \frac{3u^-^4}{-4} \\ \\ = -\frac{3}{4}* \frac{1}{u^4} \\ \\ = - \frac{3}{4u^4}$

Lembrando que ,

$u = \frac{1}{3} x-8$

Então basta substituir:

$\\ = - \frac{3}{4( \frac{1}{3}x-8)^4 }$

E por fim, colocando a constante de integração no final:

$\\ \\ = - \frac{3}{4( \frac{1}{3}x-8)^4 } +K$

Answer 2

$\sf \displaystyle \int \frac{1}{\left(\dfrac{1}{3}x-8\right)^5}\\\\\\=\int \frac{243}{\left(x-24\right)^5}dx\\\\\\=\int \frac{243}{\left(x-24\right)^5}dx\\\\\\=243\cdot \int \frac{1}{\left(x-24\right)^5}dx\\\\\\=243\cdot \int \frac{1}{u^5}du\\\\\\=243\cdot \int \:u^{-5}du\\\\\\=243\cdot \frac{u^{-5+1}}{-5+1}\\\\\\=243\cdot \frac{\left(x-24\right)^{-5+1}}{-5+1}\\\\\\=-\frac{243}{4\left(x-24\right)^4}$

$\to \boxed{\sf =-\frac{243}{4\left(x-24\right)^4}+C}$