Question

Resolva a integral:
$\int\limits^a_b { \frac{8}{1 + x^2} } \, dx$

Onde a = $\sqrt{3}$ e b = $\frac{1}{\sqrt{3} }$

Answer 1

$\int\limits^a_b { \frac{8}{1+x^2 } } \, dx$

$8\int\limits^a_b { \frac{1}{1+x^2 } } \, dx$

propriedade : $\int\limits { \frac1}{1-x^2 } } \, dx = \frac{1}{tg x}$

$8\int\limits^a_b { \frac{1}{1+x^2 } } \, dx = 8 . \frac{1}{tg(a)} - \frac{1}{tg(b)}$

$8[tan^1(a) - tang^1(b)]$

substituindo a e b

$8(tan^-^1 \sqrt{3} - tan^-^1 \frac{1}{ \sqrt{3} } )$

$8(tan^-^1 \sqrt{3} - tan^-^1 \frac{1}{ \sqrt{3} } ) = \frac{4 \pi }{3}$