Question

integral de T. Ln (2t) dt

Answer 1

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Calcular a integral indefinida:

     $\mathsf{\displaystyle\int\!t\cdot \ell n(2t)\,dt}$


Use o método de integração por partes:

     $\begin{array}{lcl} \mathsf{u=\ell n(2t)}&\quad\Rightarrow\quad &\mathsf{du=\dfrac{1}{2t}\cdot \dfrac{d}{dt}(2t)\,dt}\\\\ &&\mathsf{du=\dfrac{1}{\diagup\!\!\!\! 2t}\cdot \diagup\!\!\!\! 2\,dt}\\\\ &&\mathsf{du=\dfrac{1}{t}\,dt}\\\\\\ \mathsf{dv=t\,dt}&\quad\Leftarrow \quad&\mathsf{v=\dfrac{1}{2}\,t^2} \end{array}$


     $\mathsf{\displaystyle\int\!u\,dv=uv-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\ell n(2t)\cdot t\,dt=\ell n(2t)\cdot \frac{1}{2}\,t^2-\int\!\frac{1}{2}\,t^2\cdot \frac{1}{t}\,dt}\\\\\\ \mathsf{\displaystyle\int\!t\cdot \ell n(2t)\,dt=\frac{1}{2}\,t^2\cdot \ell n(2t)-\frac{1}{2}\int\!\frac{t^2}{t}\,dt}\\\\\\ \mathsf{\displaystyle\int\!t\cdot \ell n(2t)\,dt=\frac{1}{2}\,t^2\cdot \ell n(2t)-\frac{1}{2}\int\!t\,dt}\\\\\\ \mathsf{\displaystyle\int\!t\cdot \ell n(2t)\,dt=\frac{1}{2}\,t^2\cdot \ell n(2t)-\frac{1}{2}\cdot \left(\frac{1}{2}\,t^2\right)+C}$

     $\boxed{\begin{array}{c}\mathsf{\displaystyle\int\!t\cdot \ell n(2t)\,dt=\frac{1}{2}\,t^2\cdot \ell n(2t)-\frac{1}{4}\,t^2+C}\end{array}}$   <————   esta é a resposta.


Bons estudos! :-)