(8) $\\int e^{2x + 1}dx$

Explanation:

Step1: Let ( u = 2x + 1 )

Then ( du=2dx ), and ( dx=\frac{1}{2}du )

Step2: Substitute into the integral

[ \begin{align*} \int e^{2x + 1}dx&=\int e^{u}\cdot\frac{1}{2}du\ &=\frac{1}{2}\int e^{u}du \end{align*} ]

Step3: Integrate ( e^{u} )

Since ( \int e^{u}du=e^{u}+C ), we have ( \frac{1}{2}\int e^{u}du=\frac{1}{2}e^{u}+C )

Step4: Substitute back ( u = 2x + 1 )

( \frac{1}{2}e^{u}+C=\frac{1}{2}e^{2x + 1}+C )

Answer:

(\frac{1}{2}e^{2x + 1}+C)