Questions: Find the indefinite integral. (Remember the constant of integration.) ∫ e^(9x-1) dx

Transcript text: Find the indefinite integral. (Remember the constant of integration.) \[ \int e^{9 x-1} d x \]

Solution

Step 1: Identify \(u = ax + b\) and calculate \(du = a dx\).

Let \(u = 9x - 1\), then \(du = 9 dx\).

Step 2: Rewrite the integral in terms of \(u\).

The integral becomes \(\int k e^u \frac{du}{9}\).

Step 3: Integrate \(k e^u \frac{{1}}{{a}}\) with respect to \(u\).

The integral of \(k e^u \frac{1}{a}\) with respect to \(u\) is \(\frac{1}{9} e^u + C\).

Step 4: Substitute back \(u = ax + b\).

Substituting back \(u = 9x - 1\), we get the final answer as \(\frac{1}{9} e^{9x - 1} + C\).

The indefinite integral of \(k e^{ax + b} dx\) is \(\frac{1}{9} e^{9x - 1} + C\), rounded to 0 decimal places.

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