Question

use as regras básicas de integração para encontrar a integral indefinida

Answer 1

9) $\int{\left(\sqrt{x}+\dfrac{3}{x}-2e^{x} \right )dx}$

$\int{\left(\sqrt{x}+\dfrac{3}{x}-2e^{x} \right )dx}\\ \\ \\ =\int{x^{1/2}\,dx}+\int{\dfrac{3}{x}\,dx}+\int{-2e^{x}\,dx}\\ \\ \\ =\int{x^{1/2}\,dx}+3\int{\dfrac{dx}{x}}-2\int{e^{x}\,dx}\\ \\ \\ =\dfrac{x^{(1/2)+1}}{(1/2)+1}+3\mathrm{\,\ell n}\left|x\right|-2e^{x}+C\\ \\ \\ =\dfrac{x^{3/2}}{3/2}+3\mathrm{\,\ell n}\left|x\right|-2e^{x}+C\\ \\ \\ =\dfrac{2}{3}\,x^{3/2}+3\mathrm{\,\ell n}\left|x\right|-2e^{x}+C\\ \\ \\ =\dfrac{2}{3}\,\sqrt{x^{3}}+3\mathrm{\,\ell n}\left|x\right|-2e^{x}+C$


10) $\int{\left(\dfrac{x^{3}+2x^{2}-x}{3x} \right )dx}$

$=\int{\dfrac{\diagup\!\!\!\! x\cdot (x^{2}+2x-1)}{3\diagup\!\!\!\! x}\,dx}\\ \\ \\ =\dfrac{1}{3}\int{(x^{2}+2x-1)\,dx}\\ \\ \\ =\dfrac{1}{3}\int{x^{2}\,dx}+\dfrac{2}{3}\int{x\,dx}-\dfrac{1}{3}\int{dx}\\ \\ \\ =\dfrac{1}{3}\cdot \dfrac{x^{2+1}}{2+1}+\dfrac{2}{3}\cdot \dfrac{x^{1+1}}{1+1}-\dfrac{1}{3}\,x+C\\ \\ \\ =\dfrac{1}{3}\cdot \dfrac{x^{3}}{3}+\dfrac{\diagup\!\!\!\! 2}{3}\cdot \dfrac{x^{2}}{\diagup\!\!\!\! 2}-\dfrac{1}{3}\,x+C\\ \\ \\ =\dfrac{x^{3}}{9}+\dfrac{x^{2}}{3}-\dfrac{x}{3}+C$