# Find the integral $\int(4\;e^{3x}+1)\;dx$

$\begin{array}{1 1}\frac{4}{3}e^{3x}+x+c \\ \frac{4}{3}e^{2x}+x+c \\ \frac{3}{4}e^{3x}+x+c \\ \frac{3}{4}e^{4x}+x+c \end{array}$

 Solution: $\frac{4}{3}e^{3x} + x + k$

Toolbox:

$\int e^{nx} = \frac{1}{n} e^{nx}$

Step 1:

$\int 4. e^{3x} dx = 4. \frac{1}{3}e^3x = \frac{4}{3}e^{3x}$

Step 2:

$\int dx = x$

Step 3:

Solution = $\frac{4}{3}e^{3x} + x + k$

edited Jan 25, 2013