Question

Calcule a integral por partes

Answer 1

$I=\displaystyle\int\!\mathrm{\ell n\,}x\cdot x\,dx$


Esta é uma exceção à regra geral para escolha de $u$ e $dv.$

Para esta integral,

$\bullet\;\;u$ ficará mais simples ao ser derivada, mas

$\bullet\;\;dv$ não ficará mais simples ao ser integrado.


Observe:

$\begin{array}{lcl} u=\mathrm{\ell n\,}x&~\Rightarrow~&du=\dfrac{1}{x}\,dx\\\\ dv=x\,dx&~\Leftarrow~&v=\dfrac{x^2}{2} \end{array}$


$\displaystyle\int\!u\,dv=uv-\int\!v\,du\\\\\\ \int\!\mathrm{\ell n\,}x\cdot x\,dx=\mathrm{\ell n\,}x\cdot \frac{x^2}{2}-\int\!\frac{x^2}{2}\cdot \frac{1}{x}\,dx\\\\\\ I=\frac{1}{2}\,x^2\,\mathrm{\ell n\,}x-\frac{1}{2}\int\!\frac{x^2}{x}\,dx\\\\\\ I=\frac{1}{2}\,x^2\,\mathrm{\ell n\,}x-\frac{1}{2}\int\!x\,dx\\\\\\ I=\frac{1}{2}\,x^2\,\mathrm{\ell n\,}x-\frac{1}{2}\cdot \left(\frac{x^2}{2} \right )+C\\\\\\ I=\frac{1}{2}\,x^2\,\mathrm{\ell n\,}x-\frac{1}{4}\,x^2+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\mathrm{\ell n\,}x\cdot x\,dx=\frac{1}{2}\,x^2\,\mathrm{\ell n\,}x-\frac{1}{4}\,x^2+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \:inx\cdot x~dx\\\\\\=in\cdot \int \:xxdx\\\\\\=in\cdot \int \:x^2dx\\\\\\in\frac{x^{2+1}}{2+1}\\\\\\=\frac{inx^3}{3}\\\\\\\to \boxed{\sf =\frac{inx^3}{3}+C}$