Question

Calcule a integral por partes

Answer 1

$\displaystyle I=\int\!x\cdot 2^x\,dx$


Método de integração por partes:

$\begin{array}{lcl} u=x&~\Rightarrow~&du=dx\\\\ dv=2^x\,dx&~\Leftarrow~&v=\dfrac{2^x}{\mathrm{\ell n\,}2}\\\\ \end{array}\\\\\\ \displaystyle\int\!u\,dv=uv-\int\!v\,du\\\\\\ \int\!x\cdot 2^x\,dx=x\cdot \frac{2^x}{\mathrm{\ell n\,}2}-\int\!\frac{2^x}{\mathrm{\ell n\,}2}\,dx\\\\\\ I=\frac{x\cdot 2^x}{\mathrm{\ell n\,}2}-\frac{1}{\mathrm{\ell n\,}2}\int\!2^x\,dx\\\\\\ I=\frac{x\cdot 2^x}{\mathrm{\ell n\,}2}-\frac{1}{\mathrm{\ell n\,}2}\cdot \left(\frac{2^x}{\mathrm{\ell n\,}2} \right )+C\\\\\\ I=\frac{x\cdot 2^x}{\mathrm{\ell n\,}2}-\frac{2^x}{(\mathrm{\ell n\,}2)^2}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!x\cdot 2^x\,dx=\frac{x\cdot 2^x}{\mathrm{\ell n\,}2}-\frac{2^x}{(\mathrm{\ell n\,}2)^2}+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \:x\cdot \:2^xdx\\\\\\=\frac{2^xx}{ln \left(2\right)}-\int \frac{2^x}{ln \left(2\right)}dx\\\\\\=\frac{2^xx}{ln \left(2\right)}-\frac{2^x}{ln ^2\left(2\right)}\\\\\\\to \boxed{\sf =\frac{2^xx}{ln \left(2\right)}-\frac{2^x}{ln ^2\left(2\right)}+C}$