Question

Calcule a Integral:
∫ dx / x²+x+1

Answer 1

Resposta:

Explicação passo-a-passo:

(x  + 1/2)² = x² + x + 1/4

somando-se 3/4 em anbos os lados:

(x  + 1/2)² + 3/4 = x² + x + 1/4 + 3/4

(x  + 1/2)² + 3/4 = x² + x + 1

$\int\limits^, \frac{1}{x^2+x+1}dx=\\=\int \frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}dx = \\\\u =x + \frac{1}{2} \\\\=\int \frac{1}{u^2+\frac{3}{4}}dx = \\=\int \frac{1}{\frac{4u^2 +3}{4}}dx = \\=\int \frac{4}{4u^2 +3}dx = \\=4\int \frac{1}{4u^2 +3}dx$

Fazendo u = $\frac{\sqrt{3} }{2}$ V:

$=4\int \frac{1}{4(\frac{\sqrt{3} }{2}V)^2 +3}dx= \\=4\int \frac{1}{4(\frac{3 }{4}V^2) +3}dx=\\\\=4\int \frac{1}{{3}V^2 +3}dx=\\=4\int \frac{1}{{3}(V^2 +1)}dx=\\=\frac{4}{3} \int \frac{1}{(V^2 +1)}dx\\=\frac{4}{3}\arctan \left(v\right)$

$=\frac{4}{3}\arctan \left(\frac{2u}{\sqrt{3} } \right)\\=\frac{4}{3}\arctan \left(\frac{2(x^{2} +\frac{1}{2}) }{\sqrt{3} } \right) =\\=\frac{4}{3}\arctan \left(\frac{2x^{2} +1 }{\sqrt{3} } \right)$