\( \int \frac { \sqrt { 4 x ^ { 2 } - 9 } } { x ^ { 3 } } d x , x > \frac { 3 } { 2 } \)
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To solve the integral \( \int \frac { \sqrt { 4 x ^ { 2 } - 9 } } { x ^ { 3 } } d x \), we'll start with a substitution that simplifies the square root. We can let \( u = 4x^2 - 9 \). Then, differentiating gives us \( du = 8x \, dx \) or \( dx = \frac{du}{8x} \). However, we need to express \( x \) in terms of \( u \). From our substitution, rearranging gives us: \[ x = \frac{\sqrt{u + 9}}{2} \] Now, substituting \( x \) into our integral will help us rewrite it, but first, it's useful to express everything in terms of \( u \): Recall that: \[ \sqrt{4x^2 - 9} = \sqrt{u} \] and \[ x^3 = \left( \frac{\sqrt{u + 9}}{2} \right)^3 = \frac{(u + 9)^{3/2}}{8} \] Now rewrite the integral: \[ \int \frac{\sqrt{u}}{x^3} dx = \int \frac{\sqrt{u}}{\frac{(u + 9)^{3/2}}{8}} \cdot \frac{du}{8x} = \int \frac{8\sqrt{u}}{(u + 9)^{3/2}} \cdot \frac{du}{8x} \] As we see, this integral can become complex. Thus, it may be beneficial to revert to a trigonometric substitution instead. We can set \( x = \frac{3}{2} \sec(\theta) \), which then implies \( 4x^2 - 9 = 4 \left( \frac{3^2}{2^2} \sec^2(\theta) - \frac{9}{4} \right) = 4 \sec^2(\theta) - 9 = \tan^2(\theta)\). This yields a fresh transformation and leads to a more manageable integral where trigonometric integrals might come into play. By utilizing these methods, the integral can be computed and simplified through those trigonometric identities. Remember, as with all calculus problems, practice makes perfect, so don't be afraid to try using different substitutions or methods to really crack that integral!
