Question

integração método de substituição com fração x+1/(x-2)^2

Answer 1

$\displaystyle \int \!\dfrac{x+1}{(x-2)^2}\,dx$

Façamos a seguinte mudança de variável:

$x-2=u~~\Rightarrow~~\left\{ \!\begin{array}{l} dx=du\\\\ x=u+2 \end{array} \right.$

Substituindo, a integral fica

$=\displaystyle\int \!\dfrac{(u+2)+1}{u^2}\,du\\\\\\ =\int \!\dfrac{u+3}{u^2}\,du\\\\\\ =\int \!\left(\dfrac{u}{u^2}+\dfrac{3}{u^2} \right )du\\\\\\ =\int \!\left(\dfrac{1}{u}+\dfrac{3}{u^2} \right )du\\\\\\ =\int \!\dfrac{1}{u}\,du+\int\!\dfrac{3}{u^2}\,du\\\\\\ =\int \!\dfrac{1}{u}\,du+3\int\! u^{-2}\,du$

$=\mathrm{\ell n}|u|+3\cdot \dfrac{u^{-2+1}}{-2+1}+C\\\\\\ =\mathrm{\ell n}|u|+3\cdot \dfrac{u^{-1}}{-1}+C\\\\\\ =\mathrm{\ell n}|u|-3u^{-1}+C\\\\ =\mathrm{\ell n}|u|-\dfrac{3}{u}+C\\\\\\ =\mathrm{\ell n}|x-2|-\dfrac{3}{x-2}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle \int \!\dfrac{x+1}{(x-2)^2}\,dx=\mathrm{\ell n}|x-2|-\dfrac{3}{x-2}+C \end{array}}$