Question

algém pode resolver a integral?
$\int\limits { \frac{1}{7 x^{2} -8} } \, dx$

Answer 1

Bom, vamos fazer algumas alterações algébricas, acompanhe:

$\displaystyle \int \frac{1}{7x^2-8} \, dx \\ \\ \\ \frac{1}{8} \cdot \int \frac{1}{\displaystyle \frac{7}{8}x^2-1} \, dx \\ \\ \\ \frac{1}{8} \cdot \int \frac{1}{\displaystyle \bigg( \frac{\sqrt{7}}{\sqrt{8}} x\bigg)^2 -1} \, dx$

Fatorando o seguinte termo, obtemos:

$\displaystyle \frac{\sqrt{7}}{\sqrt{8}} \cdot \frac{\sqrt{8}}{\sqrt{8}} = \frac{\sqrt{56}}{8} = \frac{2 \sqrt{14}}{8} = \frac{\sqrt{14}}{4}$

Daí:

$\displaystyle \frac{1}{8} \cdot \int \frac{1}{\displaystyle \bigg( \frac{\sqrt{14}}{4} x \bigg)^2 -1} \, dx \\ \\ \\ u = \frac{\sqrt{14}}{4} x \\ \\ \\ du = \frac{\sqrt{14}}{4} \, dx \\ \\ \\ dx = \frac{4}{\sqrt{14}} \, du \\ \\ \\ \frac{1}{8} \cdot \frac{4}{\sqrt{14}} \cdot \int \frac{1}{u^2-1} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \int \frac{1}{u^2-1} \, du$

Podemos resolver a integral por frações parciais, veja:

$\displaystyle \frac{1}{2\sqrt{14}} \cdot \int \frac{1}{u^2-1} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \int \frac{1}{(u+1) \cdot (u-1)} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \int \frac{A}{u+1} + \frac{B}{u-1} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \int \frac{A \cdot (u-1) + B \cdot (u+1)}{(u+1) \cdot (u-1)} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \int \frac{Au-A+Bu+B}{(u+1) \cdot (u-1)} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \int \frac{(A+B) \cdot u + B-A}{(u+1) \cdot (u-1)} \, du$

Por comparação, obtemos,

$\displaystyle \frac{1}{2\sqrt{14}} \cdot \int \frac{1}{(u+1) \cdot (u-1)} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \int \frac{0u+1}{(u+1) \cdot (u-1)} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \int \frac{(A+B) \cdot u + B-A}{(u+1) \cdot (u-1)} \, du$

O seguinte sistema linear:

$\displaystyle \left \{ {{A+B=0} \atop {B-A=1}} \right.$

Multiplicando o segundo membro por -1, temos:

$\displaystyle \left \{ {{A+B=0} \atop {-B+A=-1}} \right. \\ \\ \\ 2A=-1 \\ \\ A = - \frac{1}{2}$

E portanto:

$\displaystyle A+B=0 \\ \\ \\ -\frac{1}{2}+B=0 \\ \\ \\ B = \frac{1}{2}$

Agora, temos a seguinte integral:

$\displaystyle \frac{1}{2\sqrt{14}} \cdot \int \frac{A}{u+1} + \frac{B}{u-1} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \int \frac{ \displaystyle -\frac{1}{2}}{u+1} + \frac{\displaystyle \frac{1}{2}}{u-1} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \int -\frac{1}{2 \cdot (u+1)} + \frac{1}{2 \cdot (u-1)} \, du \\ \\ \\ \frac{1}{2\sqrt{14}} \cdot \frac{1}{2} \cdot \int -\frac{1}{u+1} + \frac{1}{u-1} \, du \\ \\ \\ \frac{1}{4\sqrt{14}} \cdot \int -\frac{1}{u+1} + \frac{1}{u-1} \, du$

Fazendo a resolução, temos:

$\displaystyle \frac{1}{4\sqrt{14}} \cdot \int -\frac{1}{u+1} + \frac{1}{u-1} \, du \\ \\ \\ \frac{1}{4\sqrt{14}} \cdot \bigg( - \ln |u+1| + \ln |u-1| \bigg) + c \\ \\ \\ -\frac{1}{4\sqrt{14}} \ln |u+1| + \frac{1}{4\sqrt{14}} \ln |u-1| + c \\ \\ \\ \boxed{\boxed{ -\frac{1}{4\sqrt{14}} \ln \bigg| \frac{\sqrt{14}}{4}x+1 \bigg| + \frac{1}{4\sqrt{14}} \ln \bigg| \frac{\sqrt{14}}{4}x-1 \bigg| + c }}$