Question

calcule as integrais​

Answer 1

item a)

$\displaystyle \int [\frac{2}{\text x}-\sqrt[3]{\text x}]\text {dx} = 2.\int \frac{1}{\text x}\text{dx}-\int \text x^{\frac{1}{3}}\text{dx} \\\\\\ 2.\text{ln}|\text x| - \frac{\displaystyle \text x^{\frac{4}{3}}}{\displaystyle \frac{4}{3}}+\text C \\\\\\ \huge\boxed{2.\text{ln}|\text x| -\frac{3.\sqrt[3]{\text x^4}}{4}+\text C\ }\checkmark$

item b)

$\displaystyle \int \text{sen(x)}\text{cos}^3(\text x)\text{dx} \\\\\\ \text{Fa{\c c}amos}; \\\\ \text u=\text{cos(x)} \to \text {du}=-\text{sen(x)}\text{dx} \\\\\\ \int -\text u^3\text{du} = -\frac{\text u^4}{4}+\text C \\\\\\ \huge\boxed{\frac{-\text{cos}^4(\text x)}{4}+\text C\ } \checkmark$

item c)

$\displaystyle \int\frac{3\text x^2+2\text x}{(\text x^3+\text x^2-4)^5}\text {dx} \\\\\\ \text{Fa{\c c}amos} : \\\\ \text u = \text x^3+\text x^2-4 \to \text{du}=3\text x^2+2\text x\ \text{dx} \\\\\\ \int \frac{\text {du}}{\text u^5} \to \int \text u^{-5}\text{du} =\frac{\text u^{(-5+1)}}{-5+1}+\text C \\\\\\\ \frac{-\text u^{(-4)}}{4} +\text C \\\\\\ \huge\boxed{\frac{-1.(\text x^3+\text x^2-4)^{-4}}{4}+\text C\ }\checkmark$