18 (3 marks) (a) Differentiate $e^{2x+1}$ - HSC - SSCE Mathematics Advanced - Question 18 - 2020 - Paper 1

To find the integral, we can use integration by parts. Let:

• $u = x + 1$
• $dv = e^{2x} dx$.

Then, we differentiate and integrate:

• $du = dx$
• $v = \frac{1}{2} e^{2x}$.

Now applying the integration by parts formula:

$\int u \, dv = uv - \int v \, du,$

we have:

$\int (x + 1)e^{2x} dx = (x + 1) \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} dx.$

The integral $\int e^{2x} dx = \frac{1}{2} e^{2x}$, so:

$\int (x + 1)e^{2x} dx = (x + 1) \cdot \frac{1}{2} e^{2x} - \frac{1}{4} e^{2x} + C.$

Thus, the final answer is:

$\frac{1}{4} e^{2x} (2(x + 1) - 1) + C = \frac{1}{4} e^{2x} (2x + 1) + C.$