Question

Utilizando o método da substituição para calcular a integral ∫t √(7t² +12dt) , obetém-se:

Answer 1

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Usar o método da substituição para calcular a integral indefinida:

$\mathsf{\displaystyle\int\!t\sqrt{7t^2+12}\,dt}\\\\\\ =\mathsf{\displaystyle \int\!\frac{1}{14}\cdot 14t\sqrt{7t^2+12}\,dt}\\\\\\ =\mathsf{\displaystyle \frac{1}{14}\int\!\sqrt{7t^2+12}\cdot 14t\,dt\qquad\quad(i)}$


Substituição:

$\mathsf{7t^2+12=u\quad\Rightarrow\quad\,14t\,dt=du}$


Substituindo, a integral  (i)  fica

$=\mathsf{\displaystyle\frac{1}{14}\int\!\sqrt{u}\,du}\\\\\\ =\mathsf{\displaystyle\frac{1}{14}\int\! u^{\!\!\!\footnotesize\begin{array}{l}\mathsf{\frac{1}{2}}\!\!\!\end{array}}\,du}$


Aplicando a regra para primitivar potências,

$=\mathsf{\dfrac{1}{14}\cdot \dfrac{u^{\!\!\!\footnotesize\begin{array}{l}\mathsf{\frac{1}{2}+1} \end{array}\!\!\!}}{\frac{1}{2}+1}+C}\\\\\\ =\mathsf{\dfrac{1}{14}\cdot \dfrac{u^{\!\!\!\footnotesize\begin{array}{l}\mathsf{\frac{3}{2}} \end{array}\!\!\!}}{\frac{3}{2}}+C}\\\\\\ =\mathsf{\dfrac{1}{14}\cdot \dfrac{2}{3}\,u^{\!\!\!\footnotesize\begin{array}{l}\mathsf{\frac{3}{2}} \end{array}\!\!\!}+C}\\\\\\ =\mathsf{\dfrac{2}{42}\,u^{\!\!\!\footnotesize\begin{array}{l}\mathsf{\frac{3}{2}} \end{array}\!\!\!}+C\qquad\quad simplificando,}\\\\\\ =\mathsf{\dfrac{1}{21}\,u^{\!\!\!\footnotesize\begin{array}{l}\mathsf{\frac{3}{2}} \end{array}\!\!\!}+C}$

Substituindo de volta para a variável  t,

$=\mathsf{\dfrac{1}{21}\,(7t^2+12)^{\!\!\!\footnotesize\begin{array}{l}\mathsf{\frac{3}{2}} \end{array}\!\!\!}+C}\\\\\\\\ =\mathsf{\dfrac{1}{21}\sqrt{(7t^2+12)^3}+C}\quad\longleftarrow\quad\textsf{esta \'e a resposta.}$


Bons estudos! :-)