Question

como resolver la integral de x-1/ x(x2+2)
para resolverla por fracciones parciales

Answer 1

$Hola~~~\mathbf{Marilu} \\ \\ \int \frac{x-1}{x(x\²+2)}dx \\ \\ \underline{resoluci\acute{o}n~} \\ \\ \frac{x-1}{x(x\²+2)}= \frac{A}{x}+ \frac{Bx+C}{x\²+2} ~~--\ \textgreater \ donde~valores~de\begin{cases}A=- \frac{1}{2} \\ B= \frac{1}{2} \\ C=1 \end{cases} \\ \\Entonces: \\ \\ \int \frac{x-1}{x(x\²+2)} dx= \int\frac{A}{x}dx+\int \frac{Bx+C}{x\²+2}dx~~~--\ \textgreater \ reemplazando~valores \\ \\ ~~~~~~~~~~~~~~~~=\int \frac{- \frac{1}{2} }{x}dx+\int \frac{ \frac{1}{2}x+1 }{x\²+2} dx$

$~~~~~~~~~~~~~~~~ =-\frac{1}{2}\int \frac{1}{x}dx+\int \frac{ \frac{x+2}{2} }{x\²+2} dx \\ \\ ~~~~~~~~~~~~~~~~= -\frac{1}{2} \int \frac{1}{x}dx+ \frac{1}{2} \int \frac{x}{x\²+2} }dx+2\int \frac{1}{x\²+2}dx \\ \\ ~~~~~~~~~~~~~~~~= - \frac{1}{2}ln|x|+ \frac{1}{4}ln|x\²+2|+ \not2 \frac{ \sqrt{2} }{\not2} arctg( \frac{x }{ \sqrt{2} }) +C \\ \\ ~~~~~~~~~~~~~~~~=\boxed{- \frac{1}{2}ln|x|+ \frac{1}{4}ln|x\²+2|+ \sqrt{2}arctg( \frac{x}{ \sqrt{2}})+C}$

$\\ \mathbb{iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii}\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~te~sirva~saludos!!$