Question

dada a função: f(x)= x+1/x5, pode-se dizer que a antiderivada da função f(x) é:

Alguém consegue resolver???

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm Propriedades~b\acute asicas}\\\underline{\rm da~integral~ind~\!\!efinida}\\\blacksquare \rm1:\!\!\!\!\displaystyle\sf\int x^{n}dx=\dfrac{x^{n+1}}{n+1}+k,\forall n\ne-1\\\\\blacksquare\rm2:~\!\!\!\!\displaystyle\sf\int dx=x+k\\\\\rm\blacksquare3:\displaystyle\sf\int[\alpha f(x)\pm \beta g(x)]dx= \alpha\int f(x)dx\pm\beta\int g(x)dx\end{array}}$

$\Large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{x+1}{x^5}dx=\int(x+1)\cdot x^{-5}dx\\\displaystyle\sf\int(x^{-4}+x^{-5})dx=\int x^{-4}dx+\int x^{-5}dx\\\\\displaystyle\sf\int\dfrac{x+1}{x^5}dx=\dfrac{x^{-4+1}}{-4+1}+\dfrac{x^{-5+1}}{-5+1}+k\\\\\displaystyle\sf\int\dfrac{x+1}{x^5}dx=\dfrac{x^{-3}}{-3}+\dfrac{x^{-4}}{-4}+k\\\sf mas~\bigg(\dfrac{a}{b}\bigg)^{-n}=\bigg(\dfrac{b}{a}\bigg)^{n},com~ a\ne0\end{array}}$

$\Large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{x+1}{x^5}dx=-\dfrac{1}{3}x^{-3}-\dfrac{1}{4}x^{-4}+k\\\underline{\rm usando~a~propriedade~anterior:}\\\displaystyle\sf\int\dfrac{x+1}{x^5}dx=-\dfrac{1}{3}\bigg(\dfrac{1}{x}\bigg)^3-\dfrac{1}{4}\bigg(\dfrac{1}{x}\bigg)^4+k\\\underline{\rm por~fim:}\\\displaystyle\sf\int\dfrac{x+1}{x^5}dx=-\dfrac{1}{3x^3}-\dfrac{1}{4x^4}+k\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Resposta~garantida~por:}\\\\\huge\boxed{\boxed{\boxed{\boxed{\mathbb{C}\mathbb{Y}\mathbb{B}\mathbb{E}\mathbb{R}~\mathbb{K}\mathbb{I}\mathbb{R}\mathbb{I}\mathbb{T}\mathbb{O}}}}}\end{array}}$