Question

Resolva a: ∫√x ㏑x dx

Answer 1

$\displaystyle\int\!\sqrt{x}\,\mathrm{\ell n}\,x\,dx\\\\\\ =\int\!x^{1/2}\,\mathrm{\ell n}\,x\,dx\\\\\\ =\int\!\mathrm{\ell n}\,x\cdot x^{1/2}\,dx$


Método de integração por partes:

$\begin{array}{lcl} u=\mathrm{\ell n\,}x&~\Rightarrow~&du=\dfrac{1}{x}\,dx\\\\ dv=x^{1/2}\,dx&~\Leftarrow~&v=\dfrac{2}{3}\,x^{3/2} \end{array}\\\\\\\\ \displaystyle\int\!u\,dv=uv-\int\!v\,du\\\\\\ \!\int\!\mathrm{\ell n}\,x\cdot x^{1/2}\,dx=\mathrm{\ell n\,}x\cdot \frac{2}{3}\,x^{3/2}-\int\!\frac{2}{3}\,x^{3/2}\cdot \dfrac{1}{x}\,dx\\\\\\ \int\!x^{1/2}\,\mathrm{\ell n}\,x\,dx=\frac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\frac{2}{3}\int\!\frac{x^{3/2}}{x}\,dx$

$\displaystyle\int\!x^{1/2}\,\mathrm{\ell n}\,x\,dx=\frac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\frac{2}{3}\int\!x^{1/2}\,dx\\\\\\ \int\!x^{1/2}\,\mathrm{\ell n}\,x\,dx=\frac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\frac{2}{3}\cdot \dfrac{x^{(1/2)+1}}{\frac{1}{2}+1}+C\\\\\\ \int\!x^{1/2}\,\mathrm{\ell n}\,x\,dx=\frac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\frac{2}{3}\cdot \dfrac{x^{3/2}}{\frac{3}{2}}+C\\\\\\ \int\!x^{1/2}\,\mathrm{\ell n}\,x\,dx=\frac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\frac{2}{3}\cdot \dfrac{2}{3}\,x^{3/2}+C$


$\therefore~~\boxed{\begin{array}{c}\displaystyle\int\!\sqrt{x}\,\mathrm{\ell n}\,x\,dx=\frac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\frac{4}{9}\,x^{3/2}+C\end{array}}$


Bons estudos! :-)