1. \int \frac{d x}{3 x^{2}-2 x+4} (resp. \frac{1}{\sqrt{11}} \operatorname{arctg}\left(\frac{3 x-1}{\sqrt{11}}\right)+C )

Passo 1

O enunciado pede para calcular a integral indefinida
\int \frac{dx}{3x^{2}-2x+4}.
Aqui vamos usar o método de completar o quadrado no denominador e em seguida a fórmula básica
\int \frac{du}{u^{2}+a^{2}}=\frac{1}{a}\arctan\!\Bigl(\frac{u}{a}\Bigr)+C.

Passo 2

Primeiro, completamos o quadrado em 3x^{2}-2x+4. Fatorando 3:
3x^{2}-2x+4=3\Bigl(x^{2}-\frac{2}{3}x+\frac{4}{3}\Bigr).
Dentro do parêntese temos
x^{2}-\frac{2}{3}x+\frac{4}{3} =(x-\tfrac{1}{3})^{2}+\Bigl(\tfrac{4}{3}-\tfrac{1}{9}\Bigr) =(x-\tfrac{1}{3})^{2}+\tfrac{11}{9}.
Logo
3x^{2}-2x+4 =3\Bigl[(x-\tfrac{1}{3})^{2}+\tfrac{11}{9}\Bigr] =3\Bigl[(x-\tfrac{1}{3})^{2}+\bigl(\tfrac{\sqrt{11}}{3}\bigr)^{2}\Bigr].

Passo 3

Substituindo na integral, vem
\int \frac{dx}{3x^{2}-2x+4} =\int \frac{dx}{3\bigl[(x-\tfrac{1}{3})^{2}+(\tfrac{\sqrt{11}}{3})^{2}\bigr]} =\frac{1}{3}\int \frac{dx}{(x-\tfrac{1}{3})^{2}+(\tfrac{\sqrt{11}}{3})^{2}}.

Passo 4

Fazemos a mudança de variável u=x-\tfrac{1}{3}, logo du=dx. A integral fica
\frac{1}{3}\int \frac{du}{u^{2}+(\tfrac{\sqrt{11}}{3})^{2}}.
Aplicando a fórmula \int du/(u^{2}+a^{2})=(1/a)\arctan(u/a), com a=\tfrac{\sqrt{11}}{3}, obtemos
\frac{1}{3}\cdot\frac{1}{\tfrac{\sqrt{11}}{3}} \arctan\!\Bigl(\frac{u}{\tfrac{\sqrt{11}}{3}}\Bigr) =\frac{1}{\sqrt{11}}\arctan\!\Bigl(\frac{u}{\tfrac{\sqrt{11}}{3}}\Bigr).

Passo 5

Voltando a u=x-\tfrac{1}{3}, note que
\frac{u}{\tfrac{\sqrt{11}}{3}} =\frac{x-\tfrac{1}{3}}{\tfrac{\sqrt{11}}{3}} =\frac{3x-1}{\sqrt{11}}.
Assim a antiderivada é
\frac{1}{\sqrt{11}}\arctan\!\Bigl(\frac{3x-1}{\sqrt{11}}\Bigr)+C.

Resposta

\int \frac{dx}{3x^{2}-2x+4} =\frac{1}{\sqrt{11}}\arctan\!\Bigl(\frac{3x-1}{\sqrt{11}}\Bigr)+C.

Exercícios relacionados