# Find the following indefinite integral: {eq}\displaystyle \int \frac{\sqrt{16 - x^2}}{x^2} dx {/eq}

## Question:

Find the following indefinite integral:

{eq}\displaystyle \int \frac{\sqrt{16 - x^2}}{x^2} dx {/eq}

## Indefinite Integral:

An integral which is defined without lower limit and upper limit, then the given integral is known as the indefinite integral. The solution of an indefinite integral contains an arbitrary integration constant.

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Given

• The integral is given as {eq}\int {\frac{{\sqrt {16 - {x^2}} }}{{{x^2}}}dx} {/eq}.

Suppose {eq}x = 4\sin y {/eq}, then {eq}dx = 4\cos...