Find the value of the integral \( \int_{3}^{4} \frac{1}{x^{2} \sqrt{25-x^{2}}} \mathrm{~d} \boldsymbol{X} \). Enter your answer using "/" to represent division and "sqrt" to represent square root Be sure to use parentheses to show exactly what belongs under the square root symbol. For example, \( \frac{\sqrt{3}}{7} \) would be written as sqrt(3)/7. Present your answer as a single fraction.

We start with the integral \[ \int_{3}^{4} \frac{1}{x^{2}\sqrt{25-x^{2}}}\,dx. \] A common substitution for integrals involving \(\sqrt{25-x^{2}}\) is to set \[ x=5\sin\theta. \] Then \[ dx=5\cos\theta\,d\theta,\quad \sqrt{25-x^{2}}=\sqrt{25-25\sin^{2}\theta}=5\cos\theta. \] For the limits: - When \(x=3\), \(\sin\theta=\frac{3}{5}\) so \(\theta=\arcsin\left(\frac{3}{5}\right)\). - When \(x=4\), \(\sin\theta=\frac{4}{5}\) so \(\theta=\arcsin\left(\frac{4}{5}\right)\). Substitute everything into the integral: \[ \int_{3}^{4} \frac{1}{x^2\sqrt{25-x^2}}\,dx =\int_{\theta=\arcsin(3/5)}^{\theta=\arcsin(4/5)} \frac{1}{(5\sin\theta)^2\left(5\cos\theta\right)}\,(5\cos\theta\,d\theta). \] Simplify the integrand. Notice that the factor \(5\cos\theta\) in the numerator and denominator cancels: \[ \frac{1}{(5\sin\theta)^2\cdot 5\cos\theta}\,(5\cos\theta) = \frac{1}{25\sin^2\theta}. \] Thus the integral becomes \[ \frac{1}{25}\int_{\theta=\arcsin(3/5)}^{\theta=\arcsin(4/5)} \frac{1}{\sin^2\theta}\,d\theta =\frac{1}{25} \int_{\arcsin(3/5)}^{\arcsin(4/5)} \csc^2\theta \, d\theta. \] Recall the antiderivative of \(\csc^2\theta\) is \[ \int \csc^2\theta\,d\theta = -\cot\theta. \] Thus, \[ \frac{1}{25}\left[-\cot\theta\right]_{\theta=\arcsin(3/5)}^{\theta=\arcsin(4/5)} =\frac{1}{25}\Big[-\cot(\arcsin(4/5))+\cot(\arcsin(3/5))\Big]. \] Now, compute the cotangents: - For \(\theta=\arcsin(4/5)\): since \(\sin\theta=4/5\), then \(\cos\theta=3/5\) (because \(3^2+4^2=5^2\)). Thus, \(\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{3/5}{4/5}=\frac{3}{4}\). - For \(\theta=\arcsin(3/5)\): here, \(\sin\theta=3/5\) and \(\cos\theta=4/5\). Thus, \(\cot\theta=\frac{4/5}{3/5}=\frac{4}{3}\). Substitute these values back in: \[ \frac{1}{25}\left[\frac{4}{3}-\frac{3}{4}\right]. \] Find a common denominator for the expression within the brackets: \[ \frac{4}{3}-\frac{3}{4}=\frac{16-9}{12}=\frac{7}{12}. \] Thus, the integral equals \[ \frac{1}{25}\cdot\frac{7}{12}=\frac{7}{300}. \] In the required format, the answer is: 7/300