Question

alguem ai consegue resolver esta integral definida
$\int\limits^1_ {-1} \, \frac{xdx}{( x^{2}+1) ^{2} }$

Answer 1

Olá!

$\displaystyle \int_{-1}^{1} \frac{x \, dx}{(x^2+1)^2} \\ \\ \\ \int_{-1}^{1} \frac{x}{(x^2+1)^2} \, dx \\ \\ \\ u = x^2+1 \\ \\ du=2x \, dx \\ \\ \\ \frac{x}{2x} \cdot \int_{-1}^{1} \frac{1}{u^2} \, du$

Continuando em outra linha:

$\displaystyle \frac{1}{2} \cdot \int_{-1}^{1} u^{-2} \, du \\ \\ \\ \frac{1}{2} \cdot \frac{u^{-2+1}}{-2+1} \\ \\ \\ \frac{1}{2} \cdot \frac{u^{-1}}{-1} \\ \\ \\ -\frac{1}{2} \cdot u^{-1} \\ \\ \\ -\frac{1}{2} \cdot \frac{1}{u} \\ \\ \\ -\frac{1}{2u}$

Portanto:

$\displaystyle -\frac{1}{2 \cdot (x^2+1)} \, \, \Bigg |_{a \, = \, -1}^{b \, = \, 1}$

E de acordo com o teorema:

$\displaystyle \int_{a}^{b} f(x) \, \, dx = F(b)-F(a)$

Temos:

$\displaystyle \Bigg( -\frac{1}{2 \cdot (b^2+1)} \Bigg) - \Bigg( -\frac{1}{2 \cdot (a^2+1)} \Bigg) \\ \\ \\ \Bigg( -\frac{1}{2 \cdot (1^2+1)} \Bigg) - \Bigg( -\frac{1}{2 \cdot ((-1)^2+1)} \Bigg) \\ \\ \\ -\frac{1}{4} - \bigg(- \frac{1}{4} \bigg) \\ \\ \\ -\frac{1}{4} + \frac{1}{4} = \boxed{0 \, u.a}$