Question

Calcule a integral definida abaixo:

$\int\limits^6_3 \frac{x}{1+ x^{4} }$

Answer 1

$\displaystyle\int\limits_{3}^{6}\dfrac{x}{1+x^{4}}dx=\displaystyle\int\limits_{3}^{6}\dfrac{x}{1+(x^{2})^{2}}dx$

Vamos substituir x² por u:

$u=x^{2}~~~\therefore~~~du=2xdx~~\therefore~~~xdx=(\frac{1}{2})du$

Achando os limites de integração em u:

$x=3~~\rightarrow~~u=3^{2}=9\\x=6~~\rightarrow~~u=6^{2}=36$

Então:

$\displaystyle\int\limits_{3}^{6}\dfrac{x}{1+x^{4}}dx=\int\limits_{3}^{6}\dfrac{1}{1+(x^{2})^{2}}xdx\\\\\\\displaystyle\int\limits_{3}^{6}\dfrac{x}{1+x^{4}}dx=\int\limits_{9}^{36}\dfrac{(\frac{1}{2})}{1+u^{2}}du\\\\\\\displaystyle\int\limits_{3}^{6}\dfrac{x}{1+x^{4}}dx=\dfrac{1}{2}\int\limits_{9}^{36}\dfrac{1}{1+u^{2}}du$

Sabemos que $\dfrac{d}{du}arctg(u)=\dfrac{1}{1+u^{2}}$

Então:

$\displaystyle\int\limits_{3}^{6}\dfrac{x}{1+x^{4}}dx=\dfrac{1}{2}\int\limits_{9}^{36}\left(\dfrac{d}{du}arctg(u)\right)du$

Sabemos que $\displaystyle\int\limits_{b}^{a}\left(\dfrac{d}{dx}f(x)\right)dx=f(a)-f(b)$

Logo:

$\displaystyle\int\limits_{3}^{6}\dfrac{x}{1+x^{4}}dx=\dfrac{1}{2}[arctg(u)]_{9}^{36}\\\\\\\boxed{\boxed{\int\limits_{3}^{6}\dfrac{x}{1+x^{4}}dx=\dfrac{1}{2}\cdot(arctg(36)-arctg(9))}}$