# 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.

## Answer and Explanation: 1

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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...

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Indefinite Integrals as Anti Derivatives

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Chapter 12 / Lesson 11
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Indefinite integrals, an integral of the integrand which does not have upper or lower limits, can be used to identify individual points at specific times. Learn more about the fundamental theorem, use of antiderivatives, and indefinite integrals through examples in this lesson.