Use any method to evaluate the integral. \( \int \frac{x d x}{\sqrt{64 x^{2}-1}} \)
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To evaluate the integral \( \int \frac{x \, dx}{\sqrt{64 x^{2}-1}} \), we can use the substitution method. Let \( u = 64x^2 - 1 \). Then, we have \( du = 128x \, dx \) or \( dx = \frac{du}{128x} \). Now, we need to express \( x \) in terms of \( u \): From our substitution: \[ x^2 = \frac{u + 1}{64} \implies x = \frac{\sqrt{u + 1}}{8}. \] Next, substitute everything back into the integral. The integral becomes: \[ \int \frac{x \, dx}{\sqrt{u}} = \int \frac{\frac{\sqrt{u + 1}}{8} \cdot \frac{du}{128x}}{\sqrt{u}}. \] The \( x \) in the denominator simplifies with our expression for \( x \), and thus the integral can be transformed back into a more workable form. However, an easier approach might be direct from a standard integral that resembles our structure. Using the fact that: \[ \int \frac{x \, dx}{\sqrt{64x^2 - 1}} = \frac{1}{64} \sqrt{64x^2 - 1} + C. \] So the final answer is: \[ \frac{1}{64} \sqrt{64x^2 - 1} + C. \]
