\int \sqrt {\frac{x^2 - 9}{x}} dx


{eq}\int \sqrt {\frac{x^2 - 9}{x}} dx {/eq}

Indefinite Integral:

Integration is used to find the area under the curve and to find the volume.

It is a reverse process of differentiation.

Derivative indicates the rate of change.

It is used to find the slope of the curve.

Answer and Explanation:


{eq}I = \int \sqrt{\frac{x^2 - 9}{x}} dx \\ Let \,\,x = 3\sec{t} \\ \Rightarrow dx = 3\sec{t} \tan{t} dt \\ {/eq}

{eq}I = \int \sqrt{\frac{9\sec^{2}{t} - 9}{3\sec{t}}} 3\sec{t} \tan{t} dt \\ = \int \sqrt{\frac{9\tan^{2}{t}}{3\sec{t}}}\, 3\sec{t} \tan{t} dt\,\,\left \{ \because \sec ^2t-\tan ^2t=1 \right \} \\ = 3\sqrt{3} \int \sqrt{\sec{t}} \tan^{2}{t}\,dt {/eq}

Learn more about this topic:

Indefinite Integrals as Anti Derivatives

from Math 104: Calculus

Chapter 12 / Lesson 11

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