Question

me ajudam por favor!!!!
calculo integral

Answer 1

$\displaystyle\int{\left(\dfrac{x^{5}+2x^{2}-1}{x^{4}} \right )dx}\\ \\ \\ =\int{\left(\dfrac{\diagup\!\!\!\!\! x^{4}\cdot (x+\frac{2}{x^{2}}-\frac{1}{x^{4}})}{\diagup\!\!\!\!\! x^{4}} \right )dx}\\ \\ \\ =\int{\left(x+\dfrac{2}{x^{2}}-\dfrac{1}{x^{4}} \right )dx}\\ \\ \\ =\int{x\,dx}+2\int{\dfrac{1}{x^{2}}\,dx}-\int{\dfrac{1}{x^{4}}\,dx}\\ \\ \\ =\int{x\,dx}+2\int{x^{-2}\,dx}-\int{x^{-4}\,dx}~~~~~~\mathbf{(i)}$


Regra da integral da potência:

$\displaystyle\int{x^{n}\,dx}=\dfrac{x^{n+1}}{n+1}\,,~~(n\neq -1)$


Aplicando a regra acima em $\mathbf{(i)},$ temos

$=\dfrac{x^{1+1}}{1+1}+2\cdot \dfrac{x^{-2+1}}{-2+1}-\dfrac{x^{-4+1}}{-4+1}+C\\ \\ \\ =\dfrac{x^{2}}{2}+2\cdot \dfrac{x^{-1}}{-1}-\dfrac{x^{-3}}{-3}+C\\ \\ \\ =\dfrac{x^{2}}{2}-2x^{-1}+\dfrac{1}{3}\,x^{-3}+C\\ \\ \\ =\dfrac{x^{2}}{2}-\dfrac{2}{x}+\dfrac{1}{3x^{3}}+C$