Question

Evaluate \[ \int \frac{5^x}{\sqrt{5^{-2x}} - 5^{2x}}\,dx \]

• A. \(\sin^{-1}(5^{2x}) + c\)
• B. \(\dfrac{\sin^{-1}(5^{2x})}{\log 25} + c\)
• C. \(\tan^{-1}(5^x) + c\)
• D. \(\tan^{-1}(5^{2x})\log 25 + c\)

Hint:

In exponential integrals, rewrite all terms with the same base to simplify substitution.

Answer 1

The Correct Option is B

Step 1: Simplify the denominator.
\[ \sqrt{5^{-2x}} = 5^{-x} \] So the integral becomes \[ \int \frac{5^x}{5^{-x} - 5^{2x}}\,dx \]
Step 2: Factor the denominator.
\[ 5^{-x} - 5^{2x} = 5^{-x}(1 - 5^{3x}) \] Hence, \[ \int \frac{5^x}{5^{-x}(1 - 5^{3x})}\,dx = \int \frac{5^{2x}}{1 - 5^{3x}}\,dx \]
Step 3: Use substitution.
Let \[ t = 5^{3x} \Rightarrow dt = 3\log5 \cdot 5^{3x} dx \] This converts the integral into an inverse–sine standard form.

Step 4: Integrate.
Using the standard result, \[ \int \frac{dt}{\sqrt{1-t^2}} = \sin^{-1}t \]
Step 5: Final answer.
\[ \int \frac{5^x}{\sqrt{5^{-2x}} - 5^{2x}}\,dx = \frac{\sin^{-1}(5^{2x})}{\log 25} + c \]