Calcule a integral
\int \frac{d x}{x^{2} \sqrt{4 x^{2}-9}}

Passo 1

O enunciado pede a integral
\int \frac{dx}{x^{2}\sqrt{4x^{2}-9}}
Isso envolve uma raiz quadrada do tipo \sqrt{a^{2}x^{2}-b^{2}}, então vamos usar substituição trigonométrica.

Passo 2

Fazemos a mudança
x = \frac{3}{2}\operatorname{sec}\theta, \quad dx = \frac{3}{2}\operatorname{sec}\theta\tan\theta\,d\theta.
Note que
\sqrt{4x^{2}-9} = \sqrt{4\Bigl(\frac{9}{4}\operatorname{sec}^{2}\theta\Bigr)-9} = \sqrt{9\operatorname{sec}^{2}\theta-9} =3\tan\theta.

Passo 3

Substituindo na integral original, obtemos
\int \frac{\frac{3}{2}\operatorname{sec}\theta\tan\theta\,d\theta} {\bigl(\frac{3}{2}\operatorname{sec}\theta\bigr)^{2}\,\bigl(3\tan\theta\bigr)} =\int \frac{\tfrac{3}{2}\operatorname{sec}\theta\tan\theta} {\tfrac{9}{4}\operatorname{sec}^{2}\theta\cdot3\tan\theta}\,d\theta =\int \frac{2}{9}\cos\theta\,d\theta
pois as \operatorname{sec}\theta\tan\theta simplificam com \operatorname{sec}^{2}\theta\tan\theta. Então
\int \frac{2}{9}\cos\theta\,d\theta =\frac{2}{9}\sin\theta + C.

Passo 4

Voltando para x, usamos
\operatorname{sec}\theta=\tfrac{2x}{3} e
\sin\theta =\tan\theta\cos\theta =\frac{\sqrt{\operatorname{sec}^{2}\theta-1}}{\operatorname{sec}\theta} =\frac{\sqrt{\tfrac{4x^{2}}{9}-1}}{\tfrac{2x}{3}} =\frac{\sqrt{4x^{2}-9}}{2x}.
Logo
\frac{2}{9}\sin\theta =\frac{2}{9}\cdot\frac{\sqrt{4x^{2}-9}}{2x} =\frac{\sqrt{4x^{2}-9}}{9x}.

Resposta

\int \frac{dx}{x^{2}\sqrt{4x^{2}-9}} =\frac{\sqrt{4x^{2}-9}}{9x}+C.

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