Question

Integral Indefinida de: $\frac{ x^{2}}{ x^{2} +1}$

Answer 1

        $\displaystyle \int \dfrac{x^2}{1+x^2}dx=\int1-\dfrac{1}{1+x^2}dx=x+\arctan x+C$

Answer 2

$Ola'~~~~\mathbb{FELIPE} \\ \\ \int ( \frac{ x^{2} }{ x^{2} +1})dx \\ \\ Se~dividirmos~~~~ x^{2} ~\underline{|~~ x^{2} +1} ~=\ \textgreater \ ( x^{2} +1)+(-1) \\ \\ Substituindo~no~numerador~seria \\ \\ \int\bigg( \frac{( x^{2} +1)+(-1)}{ x^{2} +1}\bigg) dx~~--\ \textgreater \ pode~ficar~tambem~assim: \\ \\ \int ( \frac{ x^{2} +1}{ x^{2} +1})dx-\int ( \frac{1}{ x^{2} +1})dx~~--\ \textgreater \ cortamos~termos~comuns \\ \\ \int dx-\int( \frac{1}{ x^{2} +1})dx~~--\ \textgreater \ se~sabe~que~[\int \frac{1}{1+ u^{2} } =arctg(u) ]$

$x-arctgx+C \\ \\ \boxed{ \int( \frac{ x^{2} }{ x^{2} +1})dx=x-arctx+C }\\ \\ \mathbb{ttttttttttttttttttttttttttttttttttttttt} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~ter~ajudado!!$