Question

Qual é a alternativa que apresenta a resolução da integral indefinida ∫X*In x dx?

Answer 1

Calcular a integral indefinida:

     $\mathsf{\displaystyle\int x\,\ell n(x)\,dx}\\\\\\ \mathsf{=\displaystyle\int \ell n(x)\cdot x\,dx}$


Utilize o método de integração por partes:

     $\begin{array}{lcl} \mathsf{u=\ell n(x)}&\quad\Longrightarrow\quad&\mathsf{du=\dfrac{1}{x}\,dx}\\\\ \mathsf{dv=x\,dx}&\quad\Longleftarrow\quad&\mathsf{v=\displaystyle\int x\,dx}\\\\ &&\mathsf{v=\dfrac{x^2}{2}} \end{array}$


     $\mathsf{\displaystyle\int u\,dv=uv-\int v\,du}\\\\\\ \mathsf{\displaystyle\int \ell n(x)\cdot x\,dx=\ell n(x)\cdot \frac{x^2}{2}-\int \frac{x^2}{2}\cdot \frac{1}{x}\,dx}\\\\\\ \mathsf{\displaystyle\int x\,\ell n(x)\,dx=\frac{x^2}{2}\,\ell n(x)-\frac{1}{2}\int \frac{x^2}{x}\,dx}\\\\\\ \mathsf{\displaystyle\int x\,\ell n(x)\,dx=\frac{x^2}{2}\,\ell n(x)-\frac{1}{2}\int x\,dx}\\\\\\ \mathsf{\displaystyle\int x\,\ell n(x)\,dx=\frac{x^2}{2}\,\ell n(x)-\frac{1}{2}\cdot \frac{x^2}{2}+C}$

     $\mathsf{\displaystyle\int x\,\ell n(x)\,dx=\frac{x^2}{2}\,\ell n(x)-\frac{x^2}{4}+C\quad\longleftarrow\quad resposta.}$


Dúvidas? Comente.


Bons estudos! :-)