integral de 0 a x de 1/(1+x+x^2)

Solução

\int_{0}^{x} \frac{1}{1 + x + x ^{2}} d x

\frac{2}{3} arctan (\frac{2 x + 1}{3}) - \frac{π}{3 3}

Passos da solução

\int_{0}^{x} \frac{1}{1 + x + x ^{2}} d x

Completar o cuadrado

1 + x + x^{2} : (x + \frac{1}{2})^{2} + \frac{3}{4}

= \int_{0}^{x} \frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

Aplicar integração por substituição

: \int_{\frac{1}{2}}^{x + \frac{1}{2}} \frac{4}{4 u ^{2} + 3} d u

= \int_{\frac{1}{2}}^{x + \frac{1}{2}} \frac{4}{4 u ^{2} + 3} d u

Remover a constante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{\frac{1}{2}}^{x + \frac{1}{2}} \frac{1}{4 u ^{2} + 3} d u

Aplicar integração por substituição

= 4 \cdot \int_{\frac{1}{3}}^{\frac{2}{3} (x + \frac{1}{2})} \frac{1}{2 3 ( v ^{2} + 1 )} d v

Remover a constante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{2 3} \cdot \int_{\frac{1}{3}}^{\frac{2}{3} (x + \frac{1}{2})} \frac{1}{v ^{2} + 1} d v

Aplicar as regras de integração:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 4 \cdot \frac{1}{2 3} [arctan (v)]_{\frac{1}{3}}^{\frac{2}{3} (x + \frac{1}{2})}

Simplificar

4 \cdot \frac{1}{2 3} [arctan (v)]_{\frac{1}{3}}^{\frac{2}{3} (x + \frac{1}{2})} : \frac{2}{3} [arctan (v)]_{\frac{1}{3}}^{\frac{2 x + 1}{3}}

= \frac{2}{3} [arctan (v)]_{\frac{1}{3}}^{\frac{2 x + 1}{3}}

Calcular os limites:

arctan (\frac{2 x + 1}{3}) - \frac{π}{6}

= \frac{2}{3} (arctan (\frac{2 x + 1}{3}) - \frac{π}{6})

Simplificar

= \frac{2}{3} arctan (\frac{2 x + 1}{3}) - \frac{π}{3 3}

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