Question

$\int \frac{e^x}{\sqrt{4 - 2x}} dx$ is equal to:

• A. $\frac{1}{2} \cos^{-1} (e^x) + C$
• B. $\frac{1}{2} \sin^{-1} (e^x) + C$
• C. $\frac{e^x}{2} + C$
• D. $\sin^{-1} \left( \frac{e^x}{2} \right) + C$

Hint:

When dealing with integrals that contain an exponential function and a square root, try substitution to simplify the expression.

Answer 1

The Correct Option is D

To solve the integral $\int \frac{e^x}{\sqrt{4 - 2x}} dx$, we make a substitution to simplify it. Let: \[ u = \frac{e^x}{2}, \quad du = \frac{e^x}{2} dx \] Now the integral becomes: \[ \int \frac{e^x}{\sqrt{4 - 2x}} dx = \sin^{-1} \left( \frac{e^x}{2} \right) + C \] Thus, the answer is $\sin^{-1} \left( \frac{e^x}{2} \right) + C$.