[Integral]

por **dehcalegari** » Seg Set 30, 2013 18:02

Calcule

$\int_{}^{}\frac{dx}{{x}^{2}+3}$

por **young_jedi** » Seg Set 30, 2013 19:23

fazendo

$\sqrt3.\tan(\theta)=x$

$\frac{\sqrt{3}}{\cos^2(\theta)}=dx$

$\int\frac{1}{x^2+3}dx=\int\frac{1}{(\sqrt3.\tan(\theta))^2+3}.\frac{\sqrt3}{\cos^2(\theta)}d\theta$

$=\int\frac{1}{\left(\frac{3.\sin^2(\theta)}{\cos^2\theta}\right)+3}.\frac{\sqrt3}{\cos^2(\theta)}d\theta$

$=\int\frac{1}{3.\frac{(\sin^2(\theta)+\cos^2(\theta))}{\cos^2\theta}}.\frac{\sqrt3}{\cos^2(\theta)}d\theta$

$=\int\frac{1}{\frac{3}{\cos^2\theta}}.\frac{\sqrt3}{\cos^2(\theta)}d\theta$

$=\int\frac{\sqrt3}{3}d\theta$

$=\frac{\sqrt3}{3}\theta$

$=\frac{\sqrt3}{3}\arctan\left(\frac{x}{\sqrt3}\right)$

[Integral]

[Integral]

[Integral]

Re: [Integral]

Quem está online