Question

$\int\limits {\frac{x}{ x^{4}+ 9 } \, dx$ ?

Answer 1

facem schimbarea de variabila y=x

Answer 2

$\int\limits { \frac{x}{(x^2)^2+9} } \, dx$

Notam [tex]x^2=t (x^2)`dx=t`dt 2xdx=dt [/tex]
$xdx= \frac{dt}{2}$


$\int\limits { \frac{ \frac{dt}{2} }{t^2+9} } \, dx = \int\limits { \frac{dt}{2}* \frac{1}{t^2+9} } \, dx= \frac{1}{2} \int\limits { \frac{1}{t^2+9} } \, dx = \frac{1}{2} \int\limits { \frac{1}{t^2+3^2} } \, dx =[tex] \frac{1}{2} * \frac{1}{3} arctg \frac{t}{3}+c= \frac{1}{6}arctg \frac{x^2}{3} +C$ [/tex]