Question

use as regras básicas de integração para encontrar a integral indefinida

Answer 1

5) $\int{\left(\dfrac{\sqrt{3}}{2}-\dfrac{3}{2} \right )dx}$

$=\left(\dfrac{\sqrt{3}}{2}-\dfrac{3}{2} \right )\int{dx}\\ \\ \\ =\left(\dfrac{\sqrt{3}}{2}-\dfrac{3}{2} \right )x+C$


6) $\int{\dfrac{3\,dx}{(x+1)^{-2}}}$

Fazendo a seguinte substituição

$u=x+1\;\;\Rightarrow\;\;du=dx$

temos

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


7) $\int{\left(\sqrt[3]{x^{2}}-\dfrac{1}{x^{2}} \right )dx}$

$=\int{(x^{2/3}-x^{-2})\,dx}\\ \\ \\ =\int{x^{2/3}\,dx}-\int{x^{-2}\,dx}\\ \\ \\ =\dfrac{x^{(2/3)+1}}{(2/3)+1}-\dfrac{x^{-2+1}}{-2+1}+C\\ \\ \\ =\dfrac{x^{5/3}}{5/3}-\dfrac{x^{-1}}{-1}+C\\ \\ \\ =\dfrac{3}{5}x^{5/3}+x^{-1}+C\\ \\ \\ =\dfrac{3}{5}\sqrt[3]{x^{5}}+\dfrac{1}{x}+C$


8) $\int{t^{1/2}\,(t^{2}+t-1)\,dt}$

$=\int{(t^{1/2}\cdot t^{2}+t^{1/2}\cdot t-t^{1/2}\cdot 1)\,dt}\\ \\ \\ =\int{(t^{(1/2)+2}+t^{(1/2)+1}-t^{1/2})\,dt}\\ \\ \\ =\int{(t^{5/2}+t^{3/2}-t^{1/2})\,dt}\\ \\ \\ =\int{t^{5/2}\,dt}+\int{t^{3/2}\,dt}-\int{t^{1/2}\,dt}\\ \\ \\ =\dfrac{t^{(5/2)+1}}{(5/2)+1}+\dfrac{t^{(3/2)+1}}{(3/2)+1}-\dfrac{t^{(1/2)+1}}{(1/2)+1}+C\\ \\ \\ =\dfrac{t^{7/2}}{7/2}+\dfrac{t^{5/2}}{5/2}-\dfrac{t^{3/2}}{3/2}+C\\ \\ \\ =\dfrac{2}{7}\,t^{7/2}+\dfrac{2}{5}\,t^{5/2}-\dfrac{2}{3}\,t^{3/2}+C$