Question

como qualcular ln(1-x)dx usando a integração por partes detalhdamente

Answer 1

$\int{u\,dv}=uv-\int{v\,du}$


Para calcular

$\int{\mathrm{\ell n}\left(1-x \right )\,dx}$

fazemos

$\begin{array}{ll} u=\mathrm{\ell n}\left(1-x \right ),&dv=dx\\ \\ du=\dfrac{-dx}{1-x},&v=x \end{array}$


Substituindo na fórmula de integração por partes, temos

$\int{\mathrm{\ell n}\left(1-x \right )\,dx}=x\mathrm{\,\ell n}\left(1-x \right )-\int{x\cdot \dfrac{-1}{1-x}\,dx}\\ \\ \\ \int{\mathrm{\ell n}\left(1-x \right )\,dx}=x\mathrm{\,\ell n}\left(1-x \right )-\int{\dfrac{-x}{1-x}\,dx}\\ \\ \\ \int{\mathrm{\ell n}\left(1-x \right )\,dx}=x\mathrm{\,\ell n}\left(1-x \right )-\int{\dfrac{-1+1-x}{1-x}\,dx}\\ \\ \\ \int{\mathrm{\ell n}\left(1-x \right )\,dx}=x\mathrm{\,\ell n}\left(1-x \right )-\int{\dfrac{-1dx}{1-x}}-\int{\dfrac{1-x}{1-x}\,dx}\\ \\ \\ \int{\mathrm{\ell n}\left(1-x \right )\,dx}=x\mathrm{\,\ell n}\left(1-x \right )-\mathrm{\ell n}\left(1-x \right )-\int{dx}\\ \\ \\ \int{\mathrm{\ell n}\left(1-x \right )\,dx}=x\mathrm{\,\ell n}\left(1-x \right )-\mathrm{\ell n}\left(1-x \right )-x+C$