1. \int \frac{d x}{\sqrt{1+x+x^{2}}} \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots (resp. \operatorname{arcsinh}\left(\frac{2x+1}{\sqrt{3}}\right) + C)

Passo 1

O enunciado pede para calcular a integral
\int \frac{dx}{\sqrt{1 + x + x^2}}.
O conteúdo principal é completar o quadrado no radicando e usar a fórmula padrão
\int \frac{dt}{\sqrt{t^2 + a^2}} = \operatorname{arcsinh}\!\bigl(\tfrac{t}{a}\bigr) + C.

Passo 2

Vamos completar o quadrado em x^2 + x + 1. Note que
x^2 + x + 1 = \Bigl(x + \tfrac12\Bigr)^2 + 1 - \tfrac14 = \Bigl(x + \tfrac12\Bigr)^2 + \tfrac34.

Passo 3

Definimos a troca de variável
t = x + \tfrac12, \quad dt = dx.
Então a integral vira
\int \frac{dx}{\sqrt{1+x+x^2}} = \int \frac{dt}{\sqrt{t^2 + \tfrac34}} = \int \frac{dt}{\sqrt{t^2 + \bigl(\tfrac{\sqrt3}{2}\bigr)^2}}.

Passo 4

Usamos a primitiva conhecida para \int dt/\sqrt{t^2 + a^2}:
\int \frac{dt}{\sqrt{t^2 + a^2}} = \operatorname{arcsinh}\!\bigl(\tfrac{t}{a}\bigr) + C.
Aqui a = \tfrac{\sqrt3}{2}, logo
\int \frac{dt}{\sqrt{t^2 + (\tfrac{\sqrt3}{2})^2}} = \operatorname{arcsinh}\!\Bigl(\frac{t}{\tfrac{\sqrt3}{2}}\Bigr) + C.

Passo 5

Voltando a x, temos
\frac{t}{\tfrac{\sqrt3}{2}} = \frac{x + \tfrac12}{\tfrac{\sqrt3}{2}} = \frac{2x + 1}{\sqrt3}.
Portanto a integral original é
\operatorname{arcsinh}\!\Bigl(\frac{2x+1}{\sqrt3}\Bigr) + C.

Resposta

\displaystyle \operatorname{arcsinh}\!\bigl(\tfrac{2x+1}{\sqrt3}\bigr) + C

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