Question

$\int e^{-x \log 2} 2^x dx =$

• A. $\log x + C$
• B. $x + C$
• C. $\frac{1}{x} + C$
• D. $\frac{x^2}{2} + C$

Hint:

Always simplify the integrand before trying complex integration techniques like integration by parts. The identity $e^{k \ln(a)} = a^k$ is a very common tool used by examiners to hide simple constants or terms within intimidating-looking exponential expressions.

Answer 1

The Correct Option is B

To solve the integral \(\int e^{-x \log 2} 2^x \, dx\), we need to simplify the expression inside the integral.

First, observe that \(2^x\) can be expressed as \(e^{x \log 2}\). Therefore, the integral becomes:

\[ \int e^{-x \log 2} \cdot e^{x \log 2} \, dx \]

This further simplifies to:

\[ \int e^{(-x \log 2) + (x \log 2)} \, dx = \int e^{0} \, dx = \int 1 \, dx \]

The integral of 1 with respect to \(x\) is simply:

\[ x + C \]

Thus, the solution to the integral is:

The correct answer is \(x + C\).

Let's evaluate why other options do not match:

• \(\log x + C\): This is the integral of \(\frac{1}{x}\), not applicable here.
• \(\frac{1}{x} + C\): This also corresponds to an entirely different function.
• \(\frac{x^2}{2} + C\): This is the result of integrating \(x\), which does not apply in this context.

Therefore, the correct solution is \(x + C\).