Q. \[ \int 3^{x}\,dx \]

We want to compute the indefinite integral

\[
\int 3^{x}\,dx
\]

Step 1: Recall the rule for integrating an exponential of the form \(a^{x}\).

A standard result is:

\[
\int a^{x}\,dx=\frac{a^{x}}{\ln(a)}+C \quad \text{for } a>0,\ a\neq 1
\]

Here, \(a=3\), so \(\ln(3)\) will appear in the denominator.

Step 2: Substitute \(a=3\) into the formula.

\[
\int 3^{x}\,dx=\frac{3^{x}}{\ln(3)}+C
\]

Step 3: (Optional check by differentiating.)

Differentiate the result:

\[
\frac{d}{dx}\left(\frac{3^{x}}{\ln(3)}\right)=\frac{1}{\ln(3)}\cdot \frac{d}{dx}\left(3^{x}\right)
\]

Using \(\frac{d}{dx}\left(3^{x}\right)=3^{x}\ln(3)\):

\[
\frac{1}{\ln(3)}\cdot \left(3^{x}\ln(3)\right)=3^{x}
\]

This matches the integrand, so the answer is correct.

Final Answer:

\[
\int 3^{x}\,dx=\frac{3^{x}}{\ln(3)}+C
\]