Question

utilizando o método da substituição para calcular a integral,

Answer 1

Boa noite!

Solução!

$\displaystyle\int t \sqrt{ 7t^{2}+12 } dt$


$u=7t^{2} +12\\\\\\ du=14tdt\\\\\ \dfrac{du}{14}=tdt$

$\displaystyle\int \frac{ \sqrt{u} }{14} du\\\\\\\\\\ \frac{1}{14}\displaystyle\int \sqrt{u}~ du$


$\dfrac{1}{14}\bigg(u\bigg)^{ \dfrac{1}{2} }\\\\\\\\ \dfrac{1}{14} \frac{\bigg(u\bigg)^{ \dfrac{1}{2} +1}}{ \dfrac{1}{2}+1 } \\\\\\\\\ \dfrac{1}{14} \frac{\bigg(u\bigg)^{ \dfrac{3}{2}}}{ \dfrac{3}{2} } \\\\\\\\\ \dfrac{1}{14}.\dfrac{2}{3}\bigg(u\bigg)^{ \dfrac{3}{2}}\\\\\\\ \dfrac{1}{21}\bigg(u\bigg)^{ \dfrac{3}{2}}\\\\\\\ \dfrac{1}{21}\bigg(7t^{2}+12 \bigg)^{ \dfrac{3}{2}}+c\\\\\\\ \displaystyle\int t \sqrt{ 7t^{2}+12 } dt= \dfrac{1}{21}\bigg(7t^{2}+12 \bigg)^{ \dfrac{3}{2}}+c$


$\boxed{Resposta:\dfrac{1}{21}\bigg(7t^{2}+12 \bigg)^{ \dfrac{3}{2}}+c~~~~\boxed{Alternativa~~A}}$

Boa noite!
Bons estudos!

Answer 2

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$\displaystyle\sf\int t\sqrt{7t^2+12}~dt=\dfrac{1}{14}\int 14t\sqrt{7t^2+12}~dt\\\sf\underline{ fac_{\!\!,}a}\\\sf u=7t^2+12\implies du=14t~dt\\\displaystyle\dfrac{1}{14}\int 14t\sqrt{7t^2+12}~dt=\dfrac{1}{14}\int\sqrt{u}~du\\\displaystyle\sf\dfrac{1}{14}\int u^{\frac{1}{2}}~du=\dfrac{1}{\diagdown\!\!\!\!\!\!14_7}\cdot\dfrac{\diagdown\!\!\!2}{3}u^{\frac{3}{2}}+k=\dfrac{1}{21}u^{\frac{3}{2}}+k\\\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int t\sqrt{7t^2+12}~dt=\dfrac{1}{21}(7t^2+12)^{\frac{3}{2}}+k}}}}$