Question

In Exercises \(43-46\) , evaluate the integral by using a substitution prior to integration by parts. $$\int e^{\sqrt{3 x+9}} d x$$

Short answer

The integration of the given function is \[ \frac{2}{3}e^{\sqrt{3x + 9}} + C \]

Step-by-step solution

Substitute a variable for simplified function form

Let \( u = \sqrt{3x + 9}\), then the function becomes much easier to integrate - it is now \( e^u \). We also have to find \(du\) in terms of \(dx\) by differentiating \(u\) with respect to \(x\), which yields \( du = 1.5 dx\). This means that \(dx = \frac{2}{3} du \). Our integral is now transformed: \[ \int e^u \cdot \frac{2}{3} du \]

Integrate e^u

The integral of \(e^u\) is simply \(e^u\), so the original function simplifies to \[ \frac{2}{3} e^u + C \] where \(C\) is the constant of integration.

Substitute the original variable

Substitute \(u\) back into the function by replacing \(u\) with \(\sqrt{3x + 9}\). This yields the final answer: \[ \frac{2}{3}e^{\sqrt{3x + 9}} + C \]