Question

calcule a integral: integral x+1dx/(x-2)^2

Answer 1

          $\dfrac{x+1}{(x-2)^2}=\dfrac{A}{x-2}+\dfrac{B}{(x-2)^2}\\ \\ \\ \dfrac{x+1}{(x-2)^2}=\dfrac{A(x-2)+B}{(x-2)^2}\\ \\ \\ \dfrac{x+1}{(x-2)^2}=\dfrac{Ax-2A+B}{(x-2)^2}\\ \\ \\ \text{ent\~ao: }x+1=Ax+(-2A+B)\\ \\ \\ A=1\wedge B=3\\ \\ \\ \text{por ende: }\\ \\ \dfrac{x+1}{(x-2)^2}=\dfrac{1}{x-2}+\dfrac{3}{(x-2)^2}\\ \\ \\ \\ \displaystyle \int\dfrac{x+1}{(x-2)^2}\,dx=\int\dfrac{1}{x-2}+\dfrac{3}{(x-2)^2}\,dx$
 
        $\displaystyle \int\dfrac{x+1}{(x-2)^2}\,dx=\int\dfrac{1}{x-2}\,dx+\int\dfrac{3}{(x-2)^2}\,dx\\ \\ \\ \boxed{\int\dfrac{x+1}{(x-2)^2}\,dx=\ln|x-2|-\dfrac{3}{x-2}+C}$