Question

qual a integral de ln (1-x)dx

Answer 1

∫ln(1-x) dx

Mudança de variáveis

w = 1 - x
dw = - dx

∫-ln(w)dw

u = -ln(w)          dv = dw
du = -(1/w)dw    v = w

Integral por partes

∫udv = uv - ∫vdu
   
        = - w.ln(w) - ∫w.(-1/w)dw
 
       = - w.ln(w) + w
 
Voltando com a variável substituída
 
 = - (1 - x) . ln(1 - x) + (1 - x) + C
 
 = (x - 1) . ln(1 - x) - x + 1 + C

Answer 2

$I=\displaystyle\int\!\mathrm{\ell n}(1-x)\,dx$


Método de integração por partes:

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


$\displaystyle\int\!u\,dv=uv-\int\!v\,du\\\\\\ \int\!\mathrm{\ell n}(1-x)\,dx=\mathrm{\ell n}(1-x)\cdot x-\int\!x\cdot \left(-\,\frac{1}{1-x} \right)dx\\\\\\ \int\!\mathrm{\ell n}(1-x)\,dx=x\,\mathrm{\ell n}(1-x)-\int\!\frac{-x}{1-x}\,dx\\\\\\ \int\!\mathrm{\ell n}(1-x)\,dx=x\,\mathrm{\ell n}(1-x)-\int\!\frac{1-x-1}{1-x}\,dx\\\\\\ \int\!\mathrm{\ell n}(1-x)\,dx=x\,\mathrm{\ell n}(1-x)-\int\!\left(\frac{1-x}{1-x}-\frac{1}{1-x} \right )dx$

$\displaystyle\int\!\mathrm{\ell n}(1-x)\,dx=x\,\mathrm{\ell n}(1-x)-\int\!\left(1-\frac{1}{1-x} \right )dx\\\\\\ \int\!\mathrm{\ell n}(1-x)\,dx=x\,\mathrm{\ell n}(1-x)-\int\!dx+\int\!\frac{1}{1-x} \,dx\\\\\\ \int\!\mathrm{\ell n}(1-x)\,dx=x\,\mathrm{\ell n}(1-x)-x-\mathrm{\ell n}(1-x)+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\mathrm{\ell n}(1-x)\,dx=(x-1)\,\mathrm{\ell n}(1-x)-x+C \end{array}}$


Bons estudos! :-)