Question

como resolver integral de⌠ln(1-x)dx por partes

Answer 1

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Calcular a integral indefinida:

$\mathsf{\displaystyle\int\!\ell n(1-x)\,dx\qquad\quad (x>1)}$


Método de integração por partes:

$\begin{array}{lcl} \mathsf{u=\ell n(1-x)}&\quad\Rightarrow\quad&\mathsf{du=-\,\dfrac{1}{1-x}\,dx}\\\\ \mathsf{dv=dx}&\quad\Leftarrow\quad&\mathsf{v=x}\\\\ \end{array}$


$\mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\ell n(1-x)\,dx=\ell n(1-x)\cdot x-\int\!x\cdot \left(-\,\frac{1}{1-x} \right )\,dx}\\\\\\ \mathsf{\displaystyle\int\!\ell n(1-x)\,dx=x\,\ell n(1-x)-\int\!\frac{-x}{1-x}\,dx}\\\\\\ \mathsf{\displaystyle\int\!\ell n(1-x)\,dx=x\,\ell n(1-x)-\int\!\frac{1-x-1}{1-x}\,dx}\\\\\\ \mathsf{\displaystyle\int\!\ell n(1-x)\,dx=x\,\ell n(1-x)-\int\!\left(\frac{1-x}{1-x}-\frac{1}{1-x}\right)\,dx}\\\\\\\mathsf{\displaystyle\int\!\ell n(1-x)\,dx=x\,\ell n(1-x)-\int\!\left(1-\frac{1}{1-x}\right)\,dx}\\\\\\ \mathsf{\displaystyle\int\!\ell n(1-x)\,dx=x\,\ell n(1-x)-\int\!dx-\int\!\frac{1}{1-x}\cdot (-1)\,dx}\\\\\\ \mathsf{\displaystyle\int\!\ell n(1-x)\,dx=x\,\ell n(1-x)-x-\int\!\frac{1}{w}\,dw\qquad (w=1-x)}\\\\\\ \mathsf{\displaystyle\int\!\ell n(1-x)\,dx=x\,\ell n(1-x)-x-\ell n|w|+C}\\\\\\ \mathsf{\displaystyle\int\!\ell n(1-x)\,dx=x\,\ell n(1-x)-x-\ell n|1-x|+C}$


Como  1 – x > 0,  temos que  |1 – x| = 1 – x.  Portanto, a integral é

$\mathsf{\displaystyle\int\!\ell n(1-x)\,dx=x\,\ell n(1-x)-x-\ell n(1-x)+C}$    <———    esta é a resposta.


Bons estudos! :-)