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

Passo 1

O enunciado pede
\int \frac{\sqrt{9 - x^{2}}}{x}\,dx.
Vamos usar substituição trigonométrica para remover a raiz do tipo \sqrt{a^{2}-x^{2}}, tomando x=3\sin\theta.

Passo 2

Fazendo
x = 3\sin\theta,\qquad dx = 3\cos\theta\,d\theta,\qquad \sqrt{9 - x^{2}} = 3\cos\theta,
a integral vira
\int \frac{3\cos\theta}{3\sin\theta}\,(3\cos\theta\,d\theta) =3\int \frac{\cos^{2}\theta}{\sin\theta}\,d\theta =3\int \bigl(\csc\theta - \sin\theta\bigr)\,d\theta.

Passo 3

Integramos termo a termo, sabendo que
\int \csc\theta\,d\theta = -\ln\bigl|\csc\theta + \cot\theta\bigr|+C \quad\text{e}\quad \int \sin\theta\,d\theta = -\cos\theta + C.
Logo
3\int (\csc\theta - \sin\theta)\,d\theta =3\Bigl(-\ln\lvert\csc\theta+\cot\theta\rvert + \cos\theta\Bigr)+C.

Passo 4

Volta a x. Temos
\sin\theta = \frac{x}{3},\quad \cos\theta = \frac{\sqrt{9 - x^{2}}}{3},\quad \csc\theta = \frac{3}{x},\quad \cot\theta = \frac{\sqrt{9 - x^{2}}}{x}.
Então
-\ln\lvert\csc\theta+\cot\theta\rvert =-\ln\Bigl\lvert\frac{3+\sqrt{9-x^{2}}}{x}\Bigr\rvert =-\ln\lvert3+\sqrt{9-x^{2}}\rvert + \ln\lvert x\rvert
e
3\cos\theta = \sqrt{9 - x^{2}}.
Substituindo em
3\Bigl(-\ln\lvert\csc\theta+\cot\theta\rvert + \cos\theta\Bigr)
obtemos
\sqrt{9 - x^{2}} +3\ln\Bigl\lvert\frac{x}{3+\sqrt{9-x^{2}}}\Bigr\rvert + C.

Resposta

\boxed{ \int \frac{\sqrt{9 - x^{2}}}{x}\,dx =\sqrt{9 - x^{2}} +3\ln\Bigl\lvert\frac{x}{3+\sqrt{9 - x^{2}}}\Bigr\rvert + C }.

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