If \(\int\left[\left(5^{x} d x\right) / \sqrt{\left(25^{x}-1\right)}\right]=k \log \mid 5^{x}+\sqrt{\left(25^{x}-1\right) \mid+c \text { then }}\) \(\mathrm{k}=\) (a) \(\log _{\mathrm{e}}^{(1 / 5)}\) (b) \(\left[1 /\left(\log _{\mathrm{e}} 5\right)\right]\) (c) \(\log _{\mathrm{e}} 25\) (d) \(\log _{\mathrm{e}}^{(1 / 25)}\)

Given integral: \[ \int\left[\left(5^{x} d x\right) / \sqrt{\left(25^{x}-1\right)}\right]=k \log \mid 5^{x}+\sqrt{\left(25^{x}-1\right)} \mid + c \]

Let's simplify this integral by making a substitution. Let \(\displaystyle t = 5^x\). Then, \(\displaystyle dt = 5^x \log{5} \, dx\). Thus, \(\displaystyle{dx = \dfrac{dt}{5^x \log{5}}}\).

Now we can rewrite the integral using the substitution from step 2, \[\int \frac{5^x}{\sqrt{25^x-1}} \cdot \frac{dt}{5^x \log{5}}\]

The modified expression after the substitution can be simplified as follows: \[\int \frac{dt}{\log{5} \cdot \sqrt{t^2 - 1}}\]

Now, we need to integrate the given expression to find k: \[I = \int \frac{dt}{\log{5} \cdot \sqrt{t^2 - 1}}\] The integral I is known as the standard integral for \(\displaystyle{\cosh^{-1}(t)}\) (inverse hyperbolic cosine function). It is defined as follows \[ \int \frac{dt}{\sqrt{t^2 - 1}} = \cosh^{-1}(t) + C_1 \] Substitute this back into our given integral: \[I= \frac{1}{\log{5}} \cosh^{-1}(t) + C_1 \] To find k, we need to compare this result above with the given expression after performing step 2 but before integrating, which is \[k \log \mid 5^{x}+\sqrt{\left(25^{x}-1\right)} \mid = k\log{ \mid t+\sqrt{t^2-1}\mid}\]. Now substitute back the variable \(x\), \(\displaystyle t = 5^x\), into the integral: \[ I = \frac{1}{\log{5}} \cosh^{-1}(5^x) + C_1 \]. Compare with the given expression: \[\frac{1}{\log{5}} \cosh^{-1}(5^x) + C_1 = k \log \mid 5^x+\sqrt{\left(25^x-1\right) \mid+c\]

From the comparison of the expressions, the value of k is equal to the inverse of the logarithm of 5, i.e., \[ k = \frac{1}{\log{\text{e}}{5}} \]. Therefore, the value of k is: (k) = \(\displaystyle \left[\frac{1}{\log_{\mathrm{e}} 5}\right]\) (Option b)