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Use the special integration formulas (theorem 8.2) to find the indefinite integral of 16 169x² dx. (Remember to use absolute values where appropriate. Use c for the constant of integration.)

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Final answer:

The indefinite integral of 16/169x² is found to be -16/169x + C using the standard integration formula, where C is the integration constant.

Step-by-step explanation:

To find the indefinite integral of 16/169x², we need to understand and apply the special integration formulas. Since we are dealing with a form that resembles the integral of dx/x², we can use a corresponding formula from the theorems provided. The standard integration formula for 1/x² is -1/x + C, where C is the constant of integration.

Let's apply this to our specific integral:

∫(16/169x²) dx = ∫(16/169) * (1/x²) dx = (16/169) * ∫(1/x²) dx = (16/169) * (-1/x) + C = -16/169x + C

Where C is the constant of integration. No absolute value is necessary here because x² is always non-negative, and therefore the resultant 1/x term will not result in integrating over a point of discontinuity that would mandate the use of absolute values.

User Alexei Artsimovich
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