Question

Resolver integral por substituição
$\int\limits {\frac{x}{\sqrt{1-4x^2} } } \, dx$

Answer 1

Resposta:

Explicação passo-a-passo:

$\int\frac{x}{\sqrt{1-4x^2}}dx \\Chamamos u= 1-4x^2\\\frac{du}{dx} =-8x => dx= \frac{du}{-8x}\\\\Substituindo\\\\\int \frac{x}{\sqrt{u}*(-8x)}du \ podemos \ cancelar \ o \ x\\\\\frac{-1}{8}\int {\frac{1}{\sqrt{u}}du =\\\\\frac{-1}{8}\int{u^{\frac{-1}{2}}du \ agora \ a \ integral \ por \ polinomios$

$\frac{-1}{8}*\frac{u^\frac{-1}{2}+1}{\frac{-1}{2}+1}} =\\\\\frac{-1}{8}*\frac{u^{\frac{1}{2}}}{\frac{1}{2}}}=\\\\\frac{-1}{8}2*\sqrt{u}\\\\u=1-4x^2\\\\\frac{-1}{4}*\sqrt{1-4x^2}$